An analytic function is the sum of its Taylor series on every open disc about a point that lies inside its domain of analyticity, so the radius of convergence is at least the distance to the boundary; consequently the zeros of a non-constant analytic function are isolated.
What it means. Real differentiability says nothing about power series: a C∞ function need not be the sum of its Taylor series anywhere. Complex differentiability is far stronger. One complex derivative on an open set forces the function to equal its Taylor series on every disc that fits inside that set, with coefficients f(n)(z0)/n!. So an analytic function is locally a power series, its radius of convergence is controlled by nothing but the location of the singularities, and two analytic functions that agree near a single point agree everywhere their common domain is connected.
Definition(Taylor series)Original
Let f be analytic on an open set Ω⊆C and let z0∈Ω. The Taylor series of f at z0 is the power series
Tz0f(z)=n=0∑∞n!f(n)(z0)(z−z0)n,
which makes sense because an analytic function has derivatives of every order (Cauchy's integral formula). Write
d(z0)=dist(z0,∂Ω)=inf{∣z0−w∣:w∈/Ω}
for the distance from z0 to the complement of Ω, with d(z0)=+∞ when Ω=C. Thus D(z0,d(z0)) is the largest open disc centred at z0 contained in Ω.
Theorem(Taylor's theorem for analytic functions)AdaptedAdapted from Ahlfors (1979), Ch. 5, §1.2, Thm 3, pp. 179-180; Arfken, Weber & Harris (2013), §11.5, Eqs. (11.41)-(11.42), p. 493
Let f be analytic on an open set Ω and let z0∈Ω. Then
f(z)=n=0∑∞n!f(n)(z0)(z−z0)nfor every z with ∣z−z0∣<d(z0),
the series converging absolutely there and uniformly on each closed disc ∣z−z0∣≤r<d(z0). Moreover, for every ρ with 0<ρ<d(z0) the coefficients are given by the integral
n!f(n)(z0)=2πi1∮∣ζ−z0∣=ρ(ζ−z0)n+1f(ζ)dζ.
In particular the radius of convergence R of Tz0f satisfies R≥d(z0).
Arfken, Weber and Harris give the radius of convergence as the distance from the centre of the expansion to the nearest singularity of the function, for a centre interior to a region of analyticity, and add that the expansion is unique, that convergence is guaranteed strictly inside that circle, and that on the circle itself the series may converge at some points and diverge at others. That is the working rule, and the statement below is the precise form of its "nearest singularity" clause: a boundary point counts only if it genuinely obstructs continuation.
Theorem(The radius of convergence and the nearest singularity)AdaptedAdapted from Arfken, Weber & Harris (2013), §11.5, p. 494
Let f be analytic on the open set Ω, let z0∈Ω with d(z0)<∞, and let R be the radius of convergence of Tz0f. Call a boundary point w of Ω with ∣w−z0∣=d(z0)blocking for z0 if there is no ρ>0 and no analytic function on D(z0,d(z0))∪D(w,ρ) agreeing with f on D(z0,d(z0)). If at least one boundary point at distance d(z0) is blocking for z0, then
R=d(z0).Theorem(Zeros are isolated; the identity theorem)AdaptedAdapted from Ahlfors (1979), Ch. 4, §3.2, pp. 126-127
Let f be analytic on a region (a connected open set) Ω.
No zeros of infinite order. If f(z0)=0 and f(n)(z0)=0 for every n≥1, then f≡0 on Ω.
Factoring a zero. If f≡0 and f(z0)=0, there is a smallest h≥1 with f(h)(z0)=0 — the order of the zero — and f(z)=(z−z0)hg(z) with g analytic on Ω and g(z0)=0.
Isolation. Hence the zeros of an analytic f≡0 are isolated: each has a punctured neighbourhood free of zeros.
Identity theorem. If f and g are analytic on Ω and the set {z∈Ω:f(z)=g(z)} has an accumulation point in Ω, then f=g on all of Ω.
Idea of the proofOriginal
Cauchy's integral formula already writes f(z) as an integral in which z appears only through the kernel 1/(ζ−z). Everything follows from expanding that one kernel. On a circle of radius ρ about z0, and for z inside, the ratio (z−z0)/(ζ−z0) has modulus ∣z−z0∣/ρ<1, so the kernel is a geometric series in that ratio, converging uniformly in ζ. Integrating term by term turns the single integral into a power series whose coefficients are exactly the Cauchy integrals for the derivatives. Since ρ may be taken as close to d(z0) as we like, the series represents f on the whole open disc of radius d(z0).
The statement about zeros is a corollary of the same expansion. If all derivatives vanish at z0 the series is identically zero, so f vanishes on a disc; a connectedness argument then spreads this over Ω. Otherwise the first non-vanishing coefficient lets us factor out a power of z−z0, and what remains is continuous and non-zero at z0, hence non-zero nearby.
Fix ρ with 0<ρ<d(z0) and let Cρ be the circle ∣ζ−z0∣=ρ, positively oriented. The closed disc bounded by Cρ lies in Ω, so for every z with ∣z−z0∣<ρ Cauchy's integral formula (Cauchy's integral formula) gives
f(z)=2πi1∮Cρζ−zf(ζ)dζ.
Step 1: expand the kernel. Put q=q(ζ)=(z−z0)/(ζ−z0). Since ζ−z=(ζ−z0)−(z−z0)=(ζ−z0)(1−q),
ζ−z1=ζ−z01⋅1−q1=n=0∑∞(ζ−z0)n+1(z−z0)n,
by the geometric series, which is legitimate because ∣q∣=∣z−z0∣/ρ=:q0<1 for every ζ on Cρ.
Step 2: the expansion is uniform on the circle. Since f is continuous on the compact set Cρ, it is bounded there: ∣f∣≤M on Cρ for some finite M. The n-th term of the series obtained by multiplying the last display by f(ζ) therefore has modulus at most Mq0n/ρ at every point of Cρ, and ∑nMq0n/ρ converges because q0<1. By the Weierstrass criterion (Uniform convergence) the series ∑nf(ζ)(z−z0)n(ζ−z0)−n−1 converges uniformly in ζ∈Cρ to f(ζ)/(ζ−z).
Step 3: integrate term by term. Uniform convergence on a curve of finite length permits termwise integration (Integrating and differentiating series term by term), the error being at most 2πρ times the sup-norm of the tail. Hence
f(z)=n=0∑∞(2πi1∮Cρ(ζ−z0)n+1f(ζ)dζ)(z−z0)n.
Step 4: identify the coefficients. The n-th bracket is precisely the Cauchy integral formula for the n-th derivative (Cauchy's integral formula), so it equals f(n)(z0)/n!. This proves both the expansion and the integral formula for the coefficients, for every z in ∣z−z0∣<ρ.
Step 5: enlarge the disc. The coefficients do not depend on ρ. Given any z with ∣z−z0∣<d(z0), choose ρ between ∣z−z0∣ and d(z0); Step 4 applies at that z. So the expansion holds on the whole disc ∣z−z0∣<d(z0), and its radius of convergence is at least d(z0). Absolute convergence inside the disc of convergence, and uniform convergence on closed subdiscs, are properties of any power series (Power series).
Step 6: the radius is exactly d(z0) when a boundary point is blocking. Suppose, for a contradiction, that R>d(z0), and let w be a boundary point at distance d(z0) which is blocking for z0; by the hypothesis of that statement, d(z0) is finite. The sum S of the Taylor series is analytic on D(z0,R) (Power series) and, by Step 5, agrees with f on D(z0,d(z0)). Choose any finite ρ with 0<ρ≤R−d(z0) — for instance ρ=R−d(z0) when R is finite, and ρ=1 when R=∞. Every ζ∈D(w,ρ) satisfies ∣ζ−z0∣≤∣ζ−w∣+∣w−z0∣<ρ+d(z0)=R, so D(w,ρ)⊆D(z0,R) and hence S is analytic on D(z0,d(z0))∪D(w,ρ) and agrees with f on D(z0,d(z0)). That is exactly what "blocking" forbids. Therefore R=d(z0).
Step 7: zeros. Assume f(z0)=0 and all derivatives vanish at z0. By Step 5 every term of the series vanishes, so f≡0 on D(z0,d(z0)). Let E1 be the set of points of Ω at which f and all its derivatives vanish and E2 its complement in Ω. The argument just given shows E1 is open; E2 is open because f and each f(n) are continuous. As Ω is connected and E1=∅, we get E1=Ω and f≡0. That is part 1.
If instead f≡0 and f(z0)=0, part 1 forbids all derivatives from vanishing, so a least h≥1 with f(h)(z0)=0 exists. The Taylor series then starts at n=h, and
g(z)=m=0∑∞(m+h)!f(m+h)(z0)(z−z0)m
is analytic on D(z0,d(z0)) with g(z0)=f(h)(z0)/h!=0 and f(z)=(z−z0)hg(z) there. Away from z0 we may simply define g(z)=f(z)/(z−z0)h, which is analytic on Ω∖{z0} and agrees with the series on the punctured disc, so the two definitions patch to a single analytic g on Ω. That is part 2. Since g is continuous and g(z0)=0, g is non-zero on some disc about z0, where therefore the only zero of f is z0 itself: part 3.
For part 4, set F=f−g, analytic on Ω, and let z∗ be an accumulation point in Ω of the zero set of F. If F≡0 then by part 3 the zeros of F are isolated, so z∗ has a punctured neighbourhood containing no zero of F — contradicting that z∗ is an accumulation point of zeros. (Note F(z∗)=0 by continuity.) Hence F≡0. □
Remark(Where the hypotheses are used)AdaptedAdapted from Ahlfors (1979), Ch. 5, §1.2, p. 180
Analyticity on a closed disc inside Ω is what Step 1 needs. The expansion of the kernel requires ∣z−z0∣<∣ζ−z0∣ for all ζ on the contour; that is why the contour is a circle centred at z0 and why z must lie strictly inside it.
R≥d(z0), but equality needs the extra hypothesis. Ahlfors (p. 180) points out two separate things: R can be strictly larger than d(z0), and when it is, a point that lies in Ωand inside the circle of convergence need not be a point where the series sums to f. Take Ω=C∖(−∞,0] and f the principal branch of log. At z0=−1+iε with small ε>0 the distance to the boundary is ε, while R=∣z0∣≈1. For ε=1/20 the point z1=−1−iε lies in Ω and inside the disc of convergence, yet the series sums to logz1+2πi, not logz1: summing to 0.0012484+3.1915510i against f(z1)=0.0012484−3.0916343i. This is exactly a case in which no boundary point at distance d(z0) is blocking: the nearest boundary point is w=−1, and the branch of log continued across the slit from above is analytic on D(z0,ε)∪D(w,ρ) and agrees with f on D(z0,ε). So the hypothesis of the blocking-point statement fails, as it must, and the value R=∣z0∣ is not obtained from that statement at all: it comes from the coefficients directly. Since (log)(n)(z0)=(−1)n−1(n−1)!/z0n, the Taylor coefficients are an=(−1)n−1/(nz0n), so ∣an∣1/n=(n1/n∣z0∣)−1→1/∣z0∣ and Hadamard's formula gives R=∣z0∣. Note that the word "blocking" is defined only for boundary points at distance d(z0), so it does not apply to w=0 here, which is at distance ∣z0∣≈1 rather than ε.
Connectedness is used only in Step 7. The conclusion "f≡0 on Ω" fails on a disconnected Ω, where f may vanish on one component and not on another. Isolation of zeros (part 3), being local, needs no connectedness.
"Accumulation point inΩ" cannot be relaxed.f(z)=sin(π/z) is analytic on Ω=C∖{0} and vanishes at every z=1/n, a set accumulating at 0. But 0∈/Ω, and f≡0.
Example(Expanding a rational function, and reading off the radius)Original
Expand f(z)=1/(z2+1) about z0=1.
Partial fractions. Since z2+1=(z−i)(z+i),
f(z)=2i1(z−i1−z+i1).
Expand each term. For a∈{i,−i}, write z−a=(1−a)+(z−1) and factor out 1−a:
Simplify. Writing 1∓i=2e∓iπ/4 gives (1−i)−(n+1)=2−(n+1)/2ei(n+1)π/4 and (1+i)−(n+1)=2−(n+1)/2e−i(n+1)π/4, whose difference is 2−(n+1)/2⋅2isin((n+1)π/4). Dividing by 2i,
an=2(n+1)/2(−1)nsin4(n+1)π
The first five coefficients are
a0=21,a1=−21,a2=41,a3=0,a4=−81,
which match the derivatives of f at 1 directly: f(1)=1/2, f′(1)=−2z/(z2+1)21=−1/2, and so on.
Check against the theorem. The singularities of f are the poles ±i, each at distance ∣1∓i∣=2 from z0=1. So d(z0)=2 and the radius of convergence should be 2. The sine factor takes only the three values 0,±1 and ±1/2, so
In particular ∣an∣≤2−(n+1)/2 for every n, with equality exactly when n+1≡2(mod4) — the first instances being n=1 and n=5, where ∣an∣ is 1/2 and 1/8 — while ∣a0∣=1/2 is strictly below its bound 1/2=0.70711 and ∣a2∣=1/4 strictly below 0.35355. The upper bound gives limsupn∣an∣1/n≤2−1/2, and along the subsequence n=1,5,9,… the equality case gives ∣an∣1/n=2−(n+1)/(2n)→2−1/2. Hence limsupn∣an∣1/n=2−1/2 and R=2 by Hadamard's formula (Power series). The vanishing of a3 (and of every an with n≡3(mod4)) shows that individual coefficients may vanish without affecting R.
Example(A Taylor series with infinite radius)Original
Let f(z)=e−z2, entire, and expand at z0=0. Substituting w=−z2 in the exponential series,
e−z2=m=0∑∞m!(−1)mz2m,
so f(2m)(0)=(−1)m(2m)!/m! and f(2m+1)(0)=0. Since Ω=C we have d(0)=∞, and the theorem predicts R=∞; Hadamard's formula agrees, because ∣a2m∣1/2m=(m!)−1/2m→0. As a numerical check at z=2 the convergence is slow, because the terms grow to about 4m/m! before falling: the partial sum through m=13 is still only 0.01589, but through m=25 it is 0.0183156389, against e−4=0.0183156389.
Application(Why perturbation series in physics stop converging)Original
A physical quantity computed as a function of a coupling, a frequency or a temperature is often expanded in powers of that parameter, and the expansion is used far from where it was derived. Taylor's theorem says precisely how far one may go: the radius of convergence of the expansion about z0 is the distance from z0 to the nearest singularity of the analytic continuation of the quantity into the complex parameter plane — even if that singularity sits at a complex value with no physical meaning.
Two consequences used routinely:
Response functions. The polarisability or susceptibility of a damped oscillator, χ(ω)=1/(ω02−ω2−iγω), is expanded in powers of ω for low-frequency work. Its poles sit at complex ω with ∣ω∣≈ω0, so the low-frequency expansion converges only for ∣ω∣≲ω0: the series knows about the resonance before any term looks resonant.
Virial and high-temperature expansions. A singularity of the free energy at a complex value of the fugacity or of 1/T limits the expansion even when it lies off the physical axis. Finding where the series must fail is thus a question about complex singularities, not about the physical range.
The identity theorem is used just as often, in the opposite direction: if two calculations of the same analytic quantity agree on any curve segment, however short, they agree wherever both are defined. This is what licenses checking a closed-form answer against a series or an integral representation only in some convenient corner of parameter space.
A function analytic in an annulus is the sum of a unique two-sided power series there, whose coefficients are the Cauchy integrals over any circle in the annulus; in practice the coefficients are found by algebra on known expansions rather than by integration.
What it means. Taylor's theorem needs f analytic on a whole disc. If f misbehaves at the centre — a pole, an essential singularity, anything at all — the disc must be punctured, and a one-sided power series can no longer do the job. Allowing negative powers repairs this completely: on any annulus of analyticity there is exactly one two-sided series representing f. The negative-power part is the record of what goes wrong inside the hole, and reading off its first coefficient, a−1, is the whole content of the residue calculus.
for the open annulus with centre z0 and radii R1,R2. The degenerate cases R1=0 (a punctured disc) and R2=∞ are allowed.
A Laurent series about z0 is a formal two-sided series
n=−∞∑∞an(z−z0)n.
Split it into the regular part∑n≥0an(z−z0)n and the principal part∑n≥1a−n(z−z0)−n. We attach a value to the two-sided series at z only where both halves converge at z; the convention matters, because a two-sided series has no natural order of summation and only the separate convergence of the halves makes the sum independent of one. Setting w=1/(z−z0) turns the principal part into an ordinary power series in w, so it has a radius of convergence in w, say 1/R1 with R1∈[0,∞], and converges for ∣z−z0∣>R1; the regular part converges for ∣z−z0∣<R2, its own radius. The two therefore have a common domain precisely when R1<R2, and on A(z0;R1,R2) the sum is analytic, being a sum of two analytic functions (Power series).
Terminology differs: Riley, Hobson and Bence (Riley, Hobson & Bence (2006), §24.11, p. 855) call the non-negative-power half the analytic part rather than the regular part. "Principal part" is standard in both.
Theorem(Laurent's theorem)AdaptedAdapted from Ahlfors (1979), Ch. 5, §1.3, pp. 184-186 and Exercise 1, p. 186; Arfken, Weber & Harris (2013), §11.5, Eqs. (11.47)-(11.48), p. 495
Let f be analytic on the annulus A=A(z0;R1,R2). Then
f(z)=n=−∞∑∞an(z−z0)nfor every z∈A,
with coefficients
an=2πi1∮C(ζ−z0)n+1f(ζ)dζ,n∈Z,
where C may be taken to be any circle ∣ζ−z0∣=r with R1<r<R2, positively oriented, every such r giving the same number. More generally C may be any closed contour in A that winds once counterclockwise about z0.
The series converges absolutely on A and uniformly on every closed sub-annulus R1<ρ1≤∣z−z0∣≤ρ2<R2, and it is unique: if ∑nbn(z−z0)n converges to f on A then bn=an for all n.
Idea of the proofAdaptedAdapted from Ahlfors (1979), Ch. 5, §1.3, pp. 184-185 (the splitting $f = f_1 + f_2$)
Split the function, not the integral. Write f as a sum f1+f2 in which f1 is analytic on the whole inner disc ∣z−z0∣<R2 and f2 is analytic on the whole outer region ∣z−z0∣>R1, with limit 0 at infinity. Each piece is a Cauchy integral over a circle, so each is handled by expanding the one kernel 1/(ζ−z) as a geometric series — exactly as in the proof of Taylor's theorem, except that for f2 the roles of ζ−z0 and z−z0 are interchanged, because now ∣ζ−z0∣ is the smaller of the two. That interchange is what produces negative powers of z−z0. Adding the two expansions gives a two-sided series.
The split itself comes from Cauchy's formula read backwards: the Cauchy integral of f over a circle inside A defines an analytic function everywhere inside that circle, and over a circle outside it defines an analytic function everywhere outside. That the difference of the two is f is got by surrounding the point z with a third, small circle, so that the kernel's pole becomes a hole like the hole of the annulus; the several-holes form of Cauchy's theorem then relates the three integrals, and Cauchy's integral formula evaluates the small one as f(z).
Uniqueness needs no new idea: multiply any convergent representation by (z−z0)−m−1 and integrate over a circle, where all terms but one integrate to zero.
Fix z∈A and choose radii with R1<r1<∣z−z0∣<r2<R2. Let Cj denote the circle ∣ζ−z0∣=rj, positively oriented.
Step 1: a Cauchy formula for the annulus, built from the two results we have. We want to represent f(z) by integrals over C1 and C2 alone. Neither prerequisite gives that directly: Cauchy's integral formula (Cauchy's integral formula) is stated for one simple closed contour with f analytic on and inside it, and the deformation results of Cauchy's theorem need the integrand analytic between the contours, which f(ζ)/(ζ−z) is not, because of the pole at ζ=z. The remedy is to make that pole a hole of its own.
Choose ρ>0 small enough that the closed disc ∣ζ−z∣≤ρ lies in the open sub-annulus r1<∣ζ−z0∣<r2, and let σ be the circle ∣ζ−z∣=ρ, positively oriented. Put
g(ζ)=ζ−zf(ζ).
Then g is analytic on the closed region inside C2 and outside both C1 and σ: there f is analytic, because the region lies in A, and ζ=z, because σ has been removed. The two inner contours C1 and σ are mutually exterior, since the disc bounded by σ misses the disc bounded by C1. So the several-holes identity (part 3 of the deformation statement in Cauchy's theorem) applies with C=C2 and the two holes C1,σ:
∮C2g(ζ)dζ=∮C1g(ζ)dζ+∮σg(ζ)dζ.
The last integral is where f(z) comes from. On and inside σ the function f is analytic and z is an interior point, so Cauchy's integral formula (Cauchy's integral formula) gives ∮σf(ζ)(ζ−z)−1dζ=2πif(z). Substituting and dividing by 2πi,
(Our labelling has C1 inner and C2 outer. Arfken, Weber and Harris (Arfken, Weber & Harris (2013), §11.5, Fig. 11.9 and Eq. (11.43), pp. 494-495) use the opposite convention, C1 for the outer circle and C2 for the inner one; a reader comparing with that derivation should swap the labels.)
Step 2: the two pieces are analytic on larger sets, and do not depend on the radii. Each integral is a Cauchy-type integral ∮Cφ(ζ)(ζ−z)−1dζ with the continuous density φ=f on the circle, and the lemma on such integrals (Cauchy's integral formula) makes it an analytic function of z at every z off the circle. So f1 is defined and analytic on ∣z−z0∣<r2 — and also outside C2, though we do not need that — and likewise f2 is analytic at every z outside C1. Neither depends on the choice of radius: if r2′ is another admissible radius, the two integrands agree and f is analytic on the sub-annulus between C2 and C2′, which does not contain z, so Cauchy's theorem makes the two integrals equal. Hence f1 is analytic on all of ∣z−z0∣<R2 and f2 on all of ∣z−z0∣>R1. Because ∣f2(z)∣≤r1maxC1∣f∣/(∣z−z0∣−r1)→0 as ∣z∣→∞, the function f2 also has the limit 0 at infinity.
Step 3: expand the inner piece in non-negative powers. On C2 we have ∣z−z0∣<∣ζ−z0∣=r2, so the geometric expansion of the Cauchy kernel used in the proof of Taylor's theorem (Taylor series of analytic functions) applies verbatim and converges uniformly in ζ∈C2. Integrating term by term (Integrating and differentiating series term by term),
Step 4: expand the outer piece in negative powers. On C1 the inequality runs the other way: ∣ζ−z0∣=r1<∣z−z0∣. So factor out z−z0 instead of ζ−z0:
the ratio having modulus r1/∣z−z0∣<1 at every ζ∈C1, uniformly. Write M1=maxC1∣f∣, finite because f is continuous on the compact circle C1. Multiplying through by f(ζ), the terms of the resulting series are bounded in modulus on C1 by
∣z−z0∣M1(∣z−z0∣r1)m,
a convergent geometric majorant independent of ζ, so the convergence is uniform on C1 (Uniform convergence) and termwise integration is allowed:
Writing n=−m−1, the bracket is 2πi1∮C1f(ζ)(ζ−z0)−n−1dζ, which is the same expression as in Step 3 with C1 in place of C2. Adding Steps 3 and 4 therefore gives a two-sided series for f whose coefficients all have the stated form; by Step 2 the circle used may be any admissible one, since the integrand f(ζ)(ζ−z0)−n−1 is analytic on A and Cauchy's theorem makes its integral independent of the radius. Absolute convergence on A and uniform convergence on closed sub-annuli are then properties of the two underlying power series, in z−z0 and in (z−z0)−1 (Power series).
Step 5: uniqueness, which also yields the coefficient formula. Suppose f(z)=∑n∈Zcn(z−z0)n on A, convergent in the sense of the definition block, that is, with both halves convergent at each point of A. Fix m∈Z and a radius r∈(R1,R2), and let C be the circle of radius r. Since the regular part is a power series in z−z0 convergent on a disc of radius >r and the principal part is a power series in (z−z0)−1 convergent on a disc of radius >1/r, both converge uniformly on the circle ∣z−z0∣=r, which is a compact subset of each disc of convergence (Power series). Multiply by (z−z0)−m−1, which is bounded on C, and integrate over C; uniform convergence permits termwise integration (Integrating and differentiating series term by term) of each half, hence of the whole:
The right-hand side involves only f, not the chosen series, so any two representations have the same coefficients, which is uniqueness.
Finally, the coefficient formula holds over any closed contour C′ in A that encircles z0 once counterclockwise, as the statement asserts. The integrand f(ζ)(ζ−z0)−m−1 is analytic on all of A, and such a C′ can be deformed to the circle C within A without crossing the hole; integrals of a function analytic on a multiply connected region over contours deformable into one another there are equal (Cauchy's theorem). Hence the integral over C′ equals the integral over C. □
Remark(Where the hypotheses are used, and what the series does not tell you)AdaptedAdapted from Arfken, Weber & Harris (2013), §11.5, p. 495
The annulus, not just the punctured disc, is the natural domain. The expansion depends on which annulus of analyticity is chosen, not only on z0. The same f has different Laurent series about the same centre on different annuli, as the example below shows. Asking for "the Laurent series of f about z0" is ambiguous unless the annulus is named.
Analyticity on the whole annulus is used in Step 1. If f has a singularity in the region between C1 and C2, the several-holes identity no longer applies there — its hypothesis is analyticity on the whole closed region inside C2 and outside the inner contours — and the splitting fails. One would have to make that singularity a hole as well, which is exactly how the residue theorem arises.
R1=0 is the case used by the residue calculus. Then A is a punctured disc, f has an isolated singularity at z0, and a−1 is the residue (The residue theorem).
The coefficients need not be computed by integration. Arfken stresses the point (p. 495): contour integrals for an are often intractable, while partial fractions plus geometric and exponential series give the coefficients in a line. Uniqueness is what makes this legitimate — any way of producing a convergent two-sided series produces the Laurent series.
A Laurent series diverges at the centre. Arfken, Weber and Harris put it as: the series always diverges at z=z0, and possibly out as far as ∣z−z0∣=R1 (p. 495). The mechanism is the one in the definition block: the principal part is a power series in 1/(z−z0), so it cannot converge where 1/(z−z0) is too large.
Example(One function, two annuli, two different Laurent series)Original
Let
f(z)=(z−1)(z−3)1.
Its only singularities are simple poles at z=1 and z=3, so about z0=0 there are three annuli of analyticity: ∣z∣<1, 1<∣z∣<3, and ∣z∣>3. Start from the partial fractions
f(z)=21(z−31−z−11),
which is checked by combining the right-hand side over the common denominator: [(z−1)−(z−3)]/[(z−1)(z−3)]⋅21=f(z).
On 1<∣z∣<3. Here ∣z/3∣<1 and ∣1/z∣<1, so expand the first term in powers of z/3 and the second in powers of 1/z:
This series has an infinite principal part, yet f has no singularity at 0 at all: the principal part is reporting the pole at z=1, which lies in the hole of the annulus, not a singularity at the centre.
On ∣z∣>3. Now both ∣3/z∣<1 and ∣1/z∣<1, so both terms are expanded in powers of 1/z:
There is now no regular part at all, consistent with f(z)→0 as z→∞. The leading behaviour a−2=1 gives f(z)∼1/z2, as it must.
Numerical check. At z=1.5+0.4i, which lies in 1<∣z∣<3 since ∣z∣=1.5524, the closed form gives f(z)=−0.9209594171+0.4048173262i and the middle series, truncated at ∣n∣≤200, gives the same ten digits. At z=5+i, with ∣z∣=5.0990>3, both the closed form and the outer series give 0.0823529412−0.0705882353i. Evaluating the middle series at z=5+i would be meaningless: its regular part diverges there.
Example(A principal part from an essential singularity)Original
Let f(z)=z3e2/z on the punctured plane 0<∣z∣<∞. Substituting w=2/z in the exponential series,
The highest power present is z3, so f(z)/z3→1 as ∣z∣→∞.
A direct check of the residue: integrating f over ∣z∣=1 numerically gives ∮fdz=4.18879020i, and 2πi⋅2/3=4.18879020i.
Application(Impedance, multipoles and the Z-transform)Original
Laurent series are the standard language for "expand about a singular point" in the applied sciences.
Multipole expansions outside a source. Let a two-dimensional charge distribution ρ be confined to ∣z∣≤a and let Φ be its potential outside. Expanding ln∣z−z′∣ for ∣z′∣<∣z∣ gives, in ∣z∣>a,
Two things must be said carefully here. First, the ln∣z∣ term is the monopole: it carries the total charge λ. Second, it is not a term of a Laurent series, and no Laurent series can produce it: lnz is multivalued on the annulus ∣z∣>a, so Laurent's theorem does not apply to it at all. The theorem applies to the single-valued remainder, whose Laurent expansion is ∑n≥1cnz−n with no positive powers (it is bounded at infinity). In that expansion c1/z is the dipole — its coefficient is p1/(2πε0), so up to the fixed constant it is the dipole moment — while c2/z2 is the quadrupole and cn/zn the 2n-pole, with cn=pn/(2πε0n) — so the Laurent index −n names the 2n-pole, with the monopole sitting outside the series. Equivalently, write Φ=Rew with w the (multivalued) complex potential and apply Laurent's theorem to −w′, which is single-valued in ∣z∣>a and equals Ex−iEy: its expansion begins at z−1, with the monopole in that coefficient and the 2n-pole in the coefficient of z−n−1. Either way, the expansion is valid in the stated annulus and nowhere inside it, which is the precise form of the familiar warning that multipole expansions converge only outside the smallest circle enclosing the sources.
Response at low and high frequency. The impedance Z(ω) of a passive network has poles at its resonances. If the resonances are at radii ϱ1<ϱ2<⋯ from a chosen centre of expansion, then Z is analytic in each annulus ϱk<∣ω∣<ϱk+1 bounded by two consecutive resonances — not in one annulus reaching from the lowest to the highest, which would contain the intermediate ones — and it has one Laurent series in each. Its coefficients there separate the inductive (ω+1) from the capacitive (ω−1) behaviour, and it is the choice of annulus — that is, which resonances are inside the hole — which decides what the expansion means physically.
The Z-transform of a discrete signal.X(z)=∑nxnz−nis a Laurent series, and the region of convergence is an annulus. Uniqueness is what makes the inversion formula xn=(2πi)−1∮X(z)zn−1dz well posed, and the fact that the same X(z) expanded on different annuli recovers different signals (causal, anticausal) is the fact that a Laurent series belongs to its annulus, not just to its centre.
An isolated singularity is removable, a pole of finite order, or essential, according to whether the principal part of the Laurent series is empty, finite or infinite; equivalently according to whether the function stays bounded, tends to infinity, or comes arbitrarily close to every complex value.
What it means. A single analytic function can fail in only a few ways at an isolated bad point, and the Laurent series sorts them out completely. If the negative powers are absent, the point was never bad: the function extends analytically across it. If finitely many negative powers appear, the function blows up like a fixed integer power — a pole, and the familiar behaviour of a rational function. If infinitely many appear, the function behaves wildly: near such a point it comes arbitrarily close to every complex number. There is no fourth case and no fractional order. Branch points, which also look singular, are excluded here because they are not isolated singularities of a single-valued function (Branch points and branch cuts).
Definition(Isolated singularity)AdaptedAdapted from Ahlfors (1979), Ch. 4, §3.2, p. 127
Let f be analytic on a punctured disc 0<∣z−z0∣<δ. Then z0 is an isolated singularity of f. By Laurent's theorem (Laurent series) applied to that punctured disc, f has there a unique expansion
a removable singularity if a−n=0 for every n≥1 (the principal part is empty);
a pole of order h (h≥1) if a−h=0 and a−n=0 for all n>h (the principal part is a non-empty finite sum). A pole of order 1 is simple, of order 2double, and so on. The finite sum ∑n=1ha−n(z−z0)−n is the singular part of f at z0;
an essential singularity if a−n=0 for infinitely many n≥1.
The coefficient a−1, in all three cases, is the residue of f at z0, written Resz=z0f (The residue theorem).
A function analytic on a region Ω except for poles is meromorphic on Ω: every a∈Ω has a disc ∣z−a∣<δ contained in Ω on which f is either analytic throughout, or analytic on the punctured disc with a pole at a. Poles of a meromorphic function are isolated by this definition.
If f is analytic on ∣z∣>R, the point at infinity is treated as an isolated singularity of f by convention: ∞ is a removable singularity, a pole of order h, or an essential singularity of f according as z0=0 is a removable singularity, a pole of order h, or an essential singularity of
g(w)=f(1/w),0<∣w∣<1/R.
Equivalently, write the Laurent series of f on ∣z∣>R as ∑n∈Zanzn: then ∞ is removable if an=0 for all n≥1, a pole of order h if ah=0 and an=0 for n>h, and essential otherwise. For a pole at infinity the singular part is a polynomial in z.
Definition(Riemann's criterion for removability)AdaptedAdapted from Ahlfors (1979), Ch. 4, §3.1, Thm 7, p. 124
Let f be analytic on Ω′=Ω∖{z0} for a region Ω containing z0. There exists a function analytic on all of Ω agreeing with f on Ω′ if and only if
z→z0lim(z−z0)f(z)=0.
The extension, when it exists, is unique. In particular, boundedness of f near z0 is sufficient.
Definition(Characterisation of the three types by behaviour)AdaptedAdapted from Ahlfors (1979), Ch. 4, §3.2, pp. 127-129, Thm 9, p. 129
Let z0 be an isolated singularity of f.
z0 is removable ⟺f is bounded on some punctured neighbourhood of z0⟺limz→z0f(z) exists in C.
z0 is a pole ⟺limz→z0∣f(z)∣=∞. In that case the order h is the unique integer for which (z−z0)hf(z) has a removable singularity with non-zero limit, and
f(z)=(z−z0)−hφ(z)with φ analytic near z0,φ(z0)=0.
Equivalently, 1/f extends analytically with a zero of order exactly h at z0.
z0 is essential ⟺ neither of the above, i.e. ∣f∣ is unbounded near z0 but ∣f(z)∣ does not tend to ∞.
Casorati-Weierstrass. If z0 is an essential singularity, then f comes arbitrarily close to every complex value in every neighbourhood of z0: for every A∈C, every ε>0 and every δ>0 there is z with 0<∣z−z0∣<δ and ∣f(z)−A∣<ε.
No fractional orders. If for some real α one has limz→z0∣z−z0∣α∣f(z)∣=0, and for some other real α one has the limit ∞, then there is an integerh such that the limit is 0 for all α>h and ∞ for all α<h. This h is the algebraic order of f at z0: positive at a pole, negative at a zero, zero where f is analytic and non-vanishing.
Idea of the proofOriginal
Everything rests on Riemann's criterion, which turns an analytic question into a one-line limit. Its proof is a size estimate on the Laurent coefficients. Each a−n with n≥1 is an integral of f(ζ)(ζ−z0)n−1 over a circle of radius r, and the hypothesis (z−z0)f(z)→0 says precisely that ∣f∣ is at most ε/r on that circle once r is small. The two factors of r in the ML bound — one from the circumference, one from the hypothesis — leave εrn−1≤ε. Since ε is arbitrary, the whole principal part vanishes, and what remains is a power series, which converges on the full disc and supplies the missing value.
Given that, the pole case follows by inverting: if ∣f∣→∞ then 1/f is small, hence bounded, hence extends with the value 0, and a zero of an analytic function has a finite order (Taylor series of analytic functions). Taking h to be that order and inverting again gives f=(z−z0)−hφ.
Casorati-Weierstrass is then a contradiction argument in one move: if f stayed a distance ε away from some value A, then 1/(f−A) would be bounded, so A would be a removable singularity of... more precisely, f−A would have a pole or a removable singularity at z0, and so would f. That contradicts essentialness.
ProofAdaptedAdapted from Ahlfors (1979), Ch. 4, §3.1-3.2, pp. 124-129
Riemann's criterion.Necessity and uniqueness. If an analytic F on Ω agrees with f off z0, then F is continuous at z0, so (z−z0)f(z)=(z−z0)F(z)→0⋅F(z0)=0. Uniqueness holds because two continuous extensions must agree at z0 by taking limits along Ω′.
Sufficiency. Assume (z−z0)f(z)→0. Pick δ>0 with the punctured disc 0<∣z−z0∣<δ inside Ω. On that punctured disc f is analytic, so Laurent's theorem (Laurent series) gives a unique expansion f(z)=∑n∈Zan(z−z0)n with
a−n=2πi1∮∣ζ−z0∣=rf(ζ)(ζ−z0)n−1dζ(0<r<δ,n≥1).
Let ε>0. The hypothesis supplies rε∈(0,min(δ,1)) such that ∣(ζ−z0)f(ζ)∣≤ε, that is ∣f(ζ)∣≤ε/r, whenever ∣ζ−z0∣=r≤rε. For such an r the ML estimate (Contour integrals) gives, for every n≥1,
∣a−n∣≤2π1⋅2πr⋅rε⋅rn−1=εrn−1≤ε,
using r≤1. The coefficients do not depend on r, and ε was arbitrary, so a−n=0 for every n≥1: the principal part is empty. Hence f(z)=∑n≥0an(z−z0)n on the punctured disc. A power series convergent at points of modulus arbitrarily close to δ converges on the whole disc ∣z−z0∣<δ and is analytic there (Power series); call its sum F. Then F is analytic on the full disc and equals f off z0, so patching F on the disc with f on Ω∖{z0} gives the required extension.
Part 1. If z0 is removable in the Laurent sense, the series ∑n≥0an(z−z0)n converges on the full disc and defines an analytic extension, so f has a finite limit and is bounded near z0. Conversely, if f is bounded, say ∣f∣≤M for 0<∣z−z0∣<δ, then ∣(z−z0)f(z)∣≤M∣z−z0∣→0, so Riemann's criterion gives an analytic extension F; the Laurent coefficients of f are then the Taylor coefficients of F by uniqueness (Laurent series), so the principal part vanishes. Existence of a finite limit implies boundedness near z0, closing the cycle.
Part 2, pole ⇒ blow-up. If the principal part is the finite sum ending at a−h=0, then (z−z0)hf(z)=∑n≥0an−h(z−z0)n has a removable singularity with value a−h=0 at z0. Hence ∣f(z)∣=∣z−z0∣−h∣a−h+O(z−z0)∣→∞.
Part 2, blow-up ⇒ pole. Suppose ∣f(z)∣→∞ as z→z0. Then there is δ′≤δ with ∣f∣≥1, in particular f=0, on 0<∣z−z0∣<δ′. So g=1/f is analytic there and ∣g∣≤1, hence bounded; by Part 1 it extends analytically, and the extension has g(z0)=lim1/f=0. Since g≡0, this zero has a finite order h≥1 and g(z)=(z−z0)hgh(z) with gh analytic and gh(z0)=0 (Taylor series of analytic functions). Then φ:=1/gh is analytic and non-zero near z0 and
f(z)=g(z)1=(z−z0)−hφ(z).
Expanding the analytic φ in its Taylor series and multiplying by (z−z0)−h produces a Laurent series whose principal part stops at n=h with coefficient φ(z0)=0: a pole of order exactly h. The hypothesis ∣f∣→∞ was used twice, to get f=0 near z0 and to get g(z0)=0.
Part 3 is the definition of "neither", given that Parts 1 and 2 characterise the other two cases by behaviour.
No fractional orders. Suppose limz→z0∣z−z0∣α∣f(z)∣=0 for some real α, and (as the statement assumes) that ∣z−z0∣β∣f(z)∣→∞ for some other real β. The first limit is 0 for every larger exponent as well, hence for some integer m≥α. For that m, (z−z0)mf(z) satisfies Riemann's criterion — indeed ∣z−z0∣m∣f∣→0, so multiplying by one more factor still tends to 0 — and so extends analytically with value 0 at z0. The extension cannot vanish identically: that would force f≡0 on a punctured neighbourhood of z0, and then ∣z−z0∣β∣f∣≡0 for every β, contradicting the second hypothesis. So the extension has a zero of some finite order k≥1 at z0, giving (z−z0)mf(z)=(z−z0)kψ(z) with ψ(z0)=0, so f(z)=(z−z0)k−mψ(z). Put h=m−k. Then ∣z−z0∣α∣f∣→0 for α>h and →∞ for α<h, since the sign of α−h decides which power dominates and ∣ψ∣→∣ψ(z0)∣=0. In particular β<h<α, and the integer h is determined by the two families of exponents, so starting the argument from the exponent with limit ∞ produces the same h. (If no exponent gives either limit we are in the essential case, and the statement's hypothesis does not hold.)
Part 4, Casorati-Weierstrass. Let z0 be essential and suppose, for a contradiction, that there are A∈C, ε>0 and δ′>0 with
∣f(z)−A∣≥εfor 0<∣z−z0∣<δ′.
Then u=1/(f−A) is analytic on that punctured disc and ∣u∣≤1/ε, so by Part 1 it extends analytically across z0. If u(z0)=0 then f−A=1/u is bounded near z0, so by Part 1 again z0 is a removable singularity of f−A, hence of f. If u(z0)=0 then ∣1/u∣→∞, so ∣f−A∣→∞, so ∣f∣→∞, and by Part 2 z0 is a pole of f. Either way z0 is not essential — a contradiction. □
Remark(Where the hypotheses are used, and what the types look like)Original
"Isolated" is essential. The whole classification uses the existence of a punctured disc of analyticity, which is what Laurent's theorem needs. A branch point of z1/2 at the origin is not an isolated singularity of a single-valued analytic function, so it is outside this scheme; so is the accumulation of poles of 1/sin(π/z) at z=0, since every punctured neighbourhood of 0 contains poles 1/n.
Riemann's criterion looks weaker than boundedness but is not. Its hypothesis (z−z0)f(z)→0 is formally weaker than "f bounded near z0", and that is why it is the convenient thing to check: a crude estimate on f suffices, with no need to know that f is bounded. But by Part 1 the two conditions are equivalent for an isolated singularity, since each is equivalent to removability. So no analytic function on a punctured disc satisfies the criterion while being unbounded; in particular growth like log∣z−z0∣−1 is impossible for such an f, however natural it looks as a borderline case. What the limit form does buy is the version with (z−z0)mf(z) used in the "no fractional orders" argument, where boundedness of f itself is not available.
"Meromorphic" is not used identically everywhere. We follow Ahlfors (p. 128): analytic except for poles, of any order. Hassani (Hassani (2013), §12.1, p. 363) reserves the word for functions whose only singularities are simple poles, which is the class his pole-expansion theorem needs. Our statements about meromorphic functions are for the wider class.
Casorati-Weierstrass is sharp in one direction only. It says the image of every punctured neighbourhood is dense in C. It does not say the image is all of C minus at most one point; that stronger statement (Picard's theorem) is not proved here, and e1/z, which omits the value 0, shows that one exceptional value really can be omitted.
A pole at infinity means polynomial growth. From the definition at infinity, an entire function with a pole of order h at ∞ has a Laurent series on ∣z∣>0 with no negative powers (it is entire) and no powers above h, so it is a polynomial of degree h. Correspondingly ez, whose expansion has arbitrarily high powers, has an essential singularity at infinity.
Example(Classifying every singularity of one function)Original
The second factor is analytic at 0 with value −21, so zf(z)→−41: a finite non-zero limit. Hence z=0 is a simple pole, not the triple pole the factor z3 suggests, because the numerator has a double zero there. Its residue is −41: expanding 1/(z−2)=−21(1+2z+⋯),
f(z)=(2z1−24z+⋯)(−21−4z−⋯)=−4z1−81+⋯.
At z=2. The numerator 1−cos2=1.4161468 is non-zero and z3=8, so f has a simple pole with residue (1−cos2)/8=0.1770184.
At z=∞. Substituting z=1/w, g(w):=f(1/w)=w4(1−cos(1/w))/(1−2w). Now cos(1/w) has an essential singularity at w=0: its Laurent series on 0<∣w∣<∞ is ∑k≥0(−1)kw−2k/(2k)!, with a non-zero coefficient at w−2k for every k, so infinitely many negative powers. Suppose g had only finitely many negative powers on 0<∣w∣<1/2, say none below w−N. Then
1−cosw1=w41−2wg(w),
and multiplying a Laurent series with lowest power w−N by the analytic factor 1−2w and by w−4 produces a Laurent series with lowest power w−N−4: finitely many negative powers, contradicting the previous sentence. So g has infinitely many negative powers and ∞ is an essential singularity of f. There are no other singularities, since 1−cosz is entire.
Numerical check of the residue at 0. Integrating f over the circle ∣z∣=0.3 (which encloses only the singularity at 0) and dividing by 2πi gives −0.250000000, matching −41.
Example(A removable singularity, a double pole and an essential singularity side by side)Original
Consider the three functions on punctured neighbourhoods of the origin.
p(z)=(e2z−1)/z. Since e2z−1=2z+2z2+34z3+⋯, we get p(z)=2+2z+34z2+⋯: no principal part, so removable, with the extension p(0)=2. Riemann's criterion is visible directly: zp(z)=e2z−1→0.
q(z)=(1−cosz)/z4. From the same expansion of 1−cosz, q(z)=21z−2−241+7201z2−⋯. The principal part stops at z−2, so a double pole, with Res0q=0: a pole of order 2 can perfectly well have zero residue. Consistently, z2q(z)→21=0.
r(z)=e1/z. Its Laurent series ∑k≥0z−k/k! has infinitely many negative powers: essential. The behaviour matches part 3 of the statement: along the positive real axis r(x)=e1/x→∞, while along the negative real axis r(−x)=e−1/x→0, so ∣r∣ is unbounded yet does not tend to ∞. Casorati-Weierstrass is explicit here: to hit a target A=0 exactly, solve 1/z=LogA+2πik, giving solutions zk→0 as k→∞. (Hassani (Hassani (2013), §11.2, p. 342) makes the same construction for real targets; the numbers below are ours.) Taking A=1+i: LogA=21ln2+iπ/4=0.3465736+0.7853982i, and k=10 gives z10=1/(0.3465736+63.6172513i)=0.0000856313−0.0157185402i, with ∣z10∣=0.0157188; evaluating e1/z10 returns 1+i to full precision. Every neighbourhood of 0, however small, contains such a point.
Application(Poles as resonances, essential singularities as a warning)Original
The classification is not bookkeeping: in physics the type of a singularity carries the physics.
Poles of a response function are resonances and decay rates. The retarded Green's function of a damped mode, continued to complex frequency, has simple poles at ω=±ωr−iΓ/2. The real part of the pole position is the resonance frequency, the imaginary part the linewidth, and the residue the oscillator strength that appears in the sum rules. Fitting a measured lineshape is, in this language, locating a pole and measuring its residue.
Pole order counts degeneracy and multiplicity. A double pole in a scattering amplitude signals two coincident resonances or a threshold effect; the statement above says the order is always a positive integer, which is why "half a pole" is never the right description of data and why an apparently fractional power law points instead to a branch point (Branch points and branch cuts).
Removable singularities are the ones to look for in a formula. Expressions such as (e2z−1)/z above, or sin(kr)/(kr) in a scattering form factor, look singular and are not; a code that evaluates them naively will divide by zero at exactly the physically interesting point k→0. Riemann's criterion is the check that a finite limit exists, and the Taylor series of the extension is the stable way to evaluate it.
Essential singularities mark the breakdown of expansion. The essential singularity of e−1/g at g=0 is the standard model for a non-perturbative contribution: all its derivatives at g=0+ along the real axis vanish, so no perturbative order in g ever sees it, while Casorati-Weierstrass says the function takes every value arbitrarily near g=0 in the complex plane. Instanton and tunnelling contributions are detected exactly this way.
Two analytic functions that agree on a set with an accumulation point agree wherever both are defined, so an analytic function has at most one extension to any larger region; extensions are built by chains of overlapping expansions or by functional relations, as for the Gamma function.
What it means. An analytic function carries all of its information in any arbitrarily small piece of its domain: fix its values on a short arc and every other value is already determined. That rigidity has two faces. Positively, it means a function defined by one formula on a small set — a power series, an integral that converges only in a half-plane — has at most one analytic extension to a larger set, so any formula that reproduces it on an overlap is the extension. Negatively, it means the extension may depend on the route taken: continuing a germ along two different paths with the same endpoints can land on two different values, which is where multivaluedness comes from.
Definition(Function elements, direct continuation, chains)AdaptedAdapted from Ahlfors (1979), Ch. 8, §1.2, pp. 284-285
A function element is a pair (f,Ω) with Ω⊆C a region and f analytic on Ω.
Two function elements (f1,Ω1) and (f2,Ω2) are direct analytic continuations of one another if Ω1∩Ω2=∅ and f1=f2 on Ω1∩Ω2. One then says (f2,Ω2) is the direct continuation of (f1,Ω1) to Ω2; if it exists it is unique, as the identity theorem below shows.
A chain is a finite list (f1,Ω1),…,(fn,Ωn) in which consecutive elements are direct continuations of one another; (fn,Ωn) is then called an analytic continuation of (f1,Ω1). This is an equivalence relation, and an equivalence class is a global analytic function; each of its members is a branch. Nothing guarantees that the fk fit together into a single-valued function on ⋃kΩk: if Ωj meets Ωk for non-consecutive j,k, the two need not agree there, and this is exactly the phenomenon of multivaluedness.
A germ at ζ is an equivalence class of pairs (f,ζ), where f is analytic on some neighbourhood of ζ, two pairs being identified when the functions agree on some neighbourhood of ζ. A germ at ζ is the same data as a convergent power series centred at ζ.
Definition(Continuation along a path)AdaptedAdapted from Ahlfors (1979), Ch. 8, §1.4, p. 289
Let γ:[a,b]→C be a path. A continuation of a germ along γ is a family of germs gt at γ(t), one for each t∈[a,b], which varies continuously in t: each t0 has a neighbourhood I in [a,b] and a function element (f,Ω) with γ(I)⊆Ω such that gt is the germ of f at γ(t) for all t∈I. The initial germ is ga and the terminal germ gb.
Theorem(Identity theorem and uniqueness of continuation)AdaptedAdapted from Hassani (2013), §12.3, Thm 12.3.1 and Cor. 12.3.2, pp. 372-373; Ahlfors (1979), Ch. 4, §3.2, p. 127
Let S⊆C be a region and let f1,f2 be analytic on S.
Identity theorem. If f1=f2 on some non-empty open subset of S, or on a non-degenerate segment of a curve in S (one containing more than a single point), or more generally on a subset of S having an accumulation point in S, then f1=f2 throughout S. Hassani (Hassani (2013), Thm 12.3.1, p. 372) states the curve case without the word non-degenerate; it is needed, since agreement at one point says nothing, and the proof below uses it.
Local determination. Consequently an analytic function on S is completely determined by its restriction to any neighbourhood, however small, of any one point of S.
Uniqueness of continuation. If Ω⊆S are regions and f is analytic on Ω, there is at most one analytic function on S restricting to f on Ω. In particular, if f1 is analytic on S1, f2 on S2, S1∩S2=∅ and f1=f2at every point ofS1∩S2, then
f(z)={f1(z),f2(z),z∈S1,z∈S2,
is a well-defined analytic function on S1∪S2, and f1,f2 are analytic continuations of one another.
Uniqueness along a path. Two continuations of germs along the same path γ either agree at every t or differ at every t. Hence a continuation along γ is determined by its initial germ, when it exists.
Monodromy theorem. Suppose a germ at the initial point of γ can be continued along every path in a region Ω. If two paths γ1,γ2 in Ω with the same endpoints are homotopic in Ω (Connectedness and simple connectivity), the continuations of that germ along γ1 and along γ2 end at the same germ. In particular, a germ continuable along every path in a simply connected region determines a single-valued analytic function there.
Idea of the proofOriginal
Part 1 is really a statement about zeros. The difference g=f1−f2 vanishes on a set with an accumulation point in S, and the zeros of a non-zero analytic function are isolated (Taylor series of analytic functions), so g must be identically zero. Hassani gives a direct connectedness version of the same argument: the set where g vanishes cannot have a boundary inside S, because at such a boundary point g would vanish together with a whole neighbourhood's worth of nearby zeros.
Everything else is a corollary. Uniqueness of extension (part 3) is part 1 applied to two candidate extensions. Uniqueness along a path (part 4) is part 1 applied locally: the set of t where two continuations agree is both open and closed in [a,b], because agreement of germs is a local and open condition and so is disagreement.
The monodromy theorem is the only statement with genuine content beyond part 1, and its proof is a bisection. First turn the question into one about a single loop: the two continuations agree at the end if and only if continuing once around the loop "out along γ1, back along γ2" returns the germ one started with. That loop is the image of the perimeter of the homotopy rectangle. Now halve the rectangle. Going around the perimeter has the same effect as going around the two halves in succession, the difference being an out-and-back excursion that changes nothing; so if the perimeter fails, one half fails, then one of its quarters, and so on down to a point. The loops in this chain are not perimeters but perimeters with a connecting arc out and back — which is what makes "fails to return the initial germ" meaningful for each of them, since each is a loop at the same base point. Near the limit point a single function element covers an entire small rectangle, and a loop inside the domain of one element cannot change a germ at all.
ProofAdaptedAdapted from Hassani (2013), §12.3, proof of Thm 12.3.1, p. 372; Ahlfors (1979), Ch. 8, §1.4, Thm 1, p. 289 and §1.6, Thm 2, pp. 295-296
Part 1. Let g=f1−f2, analytic on the region S, and let E={z∈S:g(z)=0}. By hypothesis E has an accumulation point z∗∈S; note z∗∈E by continuity. If g≡0 on S then, by the isolation of zeros (Taylor series of analytic functions), there is a punctured disc about z∗ containing no point of E, contradicting that z∗ is an accumulation point of E. Hence g≡0. Agreement on a non-empty open subset, or on a non-degenerate curve segment, gives such an accumulation point, so those cases are included.
Part 2 is part 1 with f2 an arbitrary competitor agreeing on the neighbourhood.
Part 3. If F,G are analytic on S and both restrict to f on the non-empty open set Ω, part 1 gives F=G. For the gluing statement, f1 and f2 agree on S1∩S2 by hypothesis, so the piecewise definition is consistent; it is analytic because analyticity is local and every point of S1∪S2 has a neighbourhood inside S1 or inside S2.
Part 4. Let (gt) and (ht) be continuations along γ and let A={t:gt=ht}. A is open in [a,b]: if gt0=ht0, pick function elements (f,Ω) and (f~,Ω~) representing the two families on a neighbourhood I of t0; they agree on a neighbourhood of γ(t0), hence on the connected component of Ω∩Ω~ containing γ(t0) by part 1, hence gt=ht for t in a possibly smaller neighbourhood of t0. A is closed in [a,b]: equivalently, its complement is open, and this needs the same component argument. Suppose gt0=ht0, and take elements (f,Ω), (f~,Ω~) representing the two families on a neighbourhood I of t0 as above. Let V be the connected component of Ω∩Ω~ containing γ(t0). If f and f~ agreed on some subset of V with an accumulation point in V, part 1 applied on the region V would give f=f~ throughout V, hence gt0=ht0; so they do not, and in particular f=f~ on every neighbourhood of γ(t0) inside V. Shrinking I so that γ(I)⊆V, we get gt=ht for every t∈I: for such t the germs of f and f~ at γ(t) are germs of functions that, if equal near γ(t), would agree on a set with an accumulation point in V and hence on all of V, which we have excluded. So the complement of A is open. Since [a,b] is connected, A is ∅ or [a,b].
Part 5. Write p and q for the common initial and terminal points of γ1,γ2, let gp be the germ at p, and let H:R=[a,b]×[0,1]→Ω be the homotopy, so H(⋅,0)=γ1, H(⋅,1)=γ2, H(a,u)=p and H(b,u)=q for all u.
Step (i): a germ attached to each path in R out of the corner. For a path λ in R starting at the corner O=(a,0), the image H∘λ is a path in Ω starting at p; by hypothesis gp can be continued along it, and by part 4 the terminal germ is unique. Write T(λ) for that terminal germ. Three properties, all from part 4. (1) Concatenation: continuing along λμ is continuing along λ and then continuing the result along μ, so T composes. (2) Out and back: continuation along σσ−1 returns the germ it started from, because reversing a continuation along σ is a continuation along σ−1 and part 4 says it is the one. Hence inserting an out-and-back excursion into a path does not change T. (3) Constant pieces: a constant path changes nothing. In particular, by (1) and (2), two paths λ,μ in R from O with the same terminal point satisfy T(λ)=T(μ) if and only if T(λμ−1)=gp, the concatenation λμ−1 being a loop at O.
Step (ii): reduction to the perimeter. Let β be the bottom edge of R from O=(a,0) to (b,0) and μ the left edge from O to (a,1) followed by the top edge to (b,1); both are paths in R from O, and both end on the right edge after the constant pieces are taken into account. Their images under H are γ1 and (the constant path at p followed by) γ2, so by (3) T(β) and T(μ) are the terminal germs of the continuations along γ1 and along γ2. Let Γ=βρμ−1, where ρ is the right edge of R from (b,0) to (b,1); this is the perimeter of R traversed once from O, bottom edge first. Since H is constant on the right edge, (3) gives T(βρ)=T(β), and then the criterion at the end of Step (i), applied to βρ and μ, says that
T(Γ)=gp⟺the continuations along γ1 and γ2 end at the same germ,
which is the assertion of part 5. Suppose, for a contradiction, that T(Γ)=gp.
Step (iii): bisection, carrying the connecting arcs. Bisect R horizontally into a lower half and an upper half. Let π1 be the perimeter of the lower half, traversed from O in the sense that agrees with Γ on their shared edges; let τ be the left edge of R from O up to the lower-left corner of the upper half, and let π2=τΓ2τ−1, where Γ2 is the perimeter of the upper half traversed from that corner in the sense agreeing with Γ. Then π1π2 differs from Γ only by inserted out-and-back excursions — along the dividing edge and along τ — so T(π1π2)=T(Γ)=gp. If both T(π1)=gp and T(π2)=gp then by (1) T(π1π2)=gp; so at least one of the two loops fails to return gp. Call the corresponding half R(1) and write that loop as σ1Γ1σ1−1, where Γ1 is the perimeter of R(1) and σ1 is an arc in R from O to a corner of R(1) (the empty arc in the first case). Bisect R(1) vertically and repeat, alternating directions. This produces nested rectangles R⊃R(1)⊃R(2)⊃⋯ with diameters tending to 0, hence shrinking to a single point p∞, and loops
π(n)=σnΓnσn−1,T(π(n))=gp,
where each σn is an initial subarc of σn+1, by construction.
Step (iv): the arcs close up into one path. The arc added at stage n runs along edges of R(n−1) from a corner of R(n−1) to a corner of R(n), so its length is at most the perimeter Ln−1 of R(n−1). Bisection halves one side at each stage and alternates the direction, so Ln−1≤2−⌊(n−1)/2⌋L with L the perimeter of R, and
n≥1∑Ln−1≤Ln≥1∑2−⌊(n−1)/2⌋=4L<∞.
The added lengths are therefore summable: the increasing arcs σn fit together into a single rectifiable arc σ∞ in R of finite total length Λ, which we parametrise on [0,1] proportionally to arc length, so that σn is σ∞ restricted to [0,tn] with tn=(length of σn)/Λ↑1. The parametrisation is continuous at s=1 with value p∞: for s>tn the point σ∞(s) lies in R(n), whose diameter tends to 0 and whose intersection is {p∞}. So σ∞ is a path in R from O to p∞.
Step (v): the contradiction. Continue gp along H∘σ∞ and let (gs)s∈[0,1] be the resulting family. By the definition of continuation along a path there is a neighbourhood I of s=1 and a function element (f∞,Ω∞) with H(σ∞(I))⊆Ω∞ and gs the germ of f∞ at H(σ∞(s)) for s∈I. Restricting a continuation to an initial subinterval is a continuation along that subpath, so by part 4, T(σn)=gtn; and for n large enough that tn∈I, this is the germ of f∞ at H(σ∞(tn)).
Now H(p∞)∈Ω∞ and Ω∞ is open, so some disc D about H(p∞) lies in Ω∞; since H is uniformly continuous on the compact R and diamR(n)→0, we have H(R(n))⊆D⊆Ω∞ for all large n. Fix an n large enough for both conditions. Then the family "germ of f∞ at the current point" along H∘Γn is a continuation, it starts at T(σn), and Γn is a loop, so it ends at the same germ: T(σnΓn)=T(σn). By (2), T(π(n))=T(σnΓnσn−1)=gp, contradicting Step (iii).
Hence T(Γ)=gp and part 5 holds. The final sentence follows because in a simply connected region all paths with the same endpoints are homotopic. □
Remark(Where the hypotheses are used)AdaptedAdapted from Ahlfors (1979), Ch. 8, §1.4, p. 289 (continuation along some paths but not others, the shrinking radius of convergence, singular paths); Hassani (2013), §12.3, footnote 3, p. 372
Connectedness of S. Part 1 fails without it: on S=D(0,1)∪D(3,1) the function equal to 0 on the first disc and 1 on the second is analytic and agrees with 0 on an open set, but not everywhere.
The accumulation point must lie inS.f1(z)=sin(π/z) and f2=0 are analytic on S=C∖{0} and agree at every z=1/n; those points accumulate only at 0∈/S, and indeed f1=f2.
Agreement on the whole overlap in part 3, and why connectedness is not an extra hypothesis. The glue works because f1=f2 at every point of S1∩S2; no connectedness of the overlap is needed, and none is assumed. What connectedness decides is a different question: whether checking agreement somewhere in the overlap is enough. If S1∩S2 is connected, agreement on any subset of it with an accumulation point propagates to the whole overlap by part 1, so a local check suffices. If it is disconnected, it does not. The standard illustration: let S1 be the plane slit along the negative imaginary axis and S2 the plane slit along the positive imaginary axis, with f1,f2 branches of log. Their overlap is the union of the left and right open half-planes, and the branches that agree throughout the right half-plane differ by 2πi on the left — so they do not agree on the whole overlap, part 3 does not apply, and indeed the piecewise formula is not well defined.
Homotopy, not just equal endpoints, in part 5. Continuing the germ of log at z=1 along the upper and lower halves of the unit circle to z=−1 gives germs differing by 2πi: those two paths have the same endpoints but are not homotopic in C∖{0}, which is where log can be continued. A hypothesis we have kept implicit appears in Hassani (Hassani (2013), §12.3, footnote 3, p. 372), attached to the opening discussion of his section rather than to Thm 12.3.1: the region S must not be discrete (countable). For us that is automatic, since a region is by definition a non-empty connected open set.
Continuation need not exist at all.Ahlfors (Ahlfors (1979), Ch. 8, §1.4, p. 289) notes that a germ may be continuable along some paths and not others; the radius of convergence of the germ at γ(t) then tends to 0 as t approaches the obstruction, and the path is called singular.
Radii and discs. By Hadamard's formula (Power series), f1 has radius 2 and f2 radius 3, so f1 is defined on D(1,2) and f2 on D(4,3). The two discs overlap in a lens-shaped region (both contain, for instance, z=2.5), which is connected.
Sum each series. Both are geometric. For f1, with ratio −(z−1)/2,
f1(z)=21⋅1+2z−11=z+11,
and for f2, with ratio −(z−4)/3,
f2(z)=−31⋅1+3z−41=−z−11.
They do not agree. At z=2.5: f1=1/3.5=0.2857143 while f2=−1/1.5=−0.6666667. So these two function elements are not continuations of one another, despite having overlapping discs — a reminder that overlap is not enough; agreement on the overlap is the condition.
A pair that does agree. Replace f2 by the expansion of the same 1/(z+1) about the centre c=1+2i:
of centre 1+2i and radius ∣2+2i∣=22=2.8284 (the distance from the centre to the singularity at −1).
The centre must be moved off the real axis. Every disc of convergence of an expansion of 1/(z+1) has −1 on its boundary. If two such centres and −1 are collinear with the centres on the same side — for instance the centres 1 and 4, radii 2 and 5 — the two discs are internally tangent at −1 and the larger simply contains the smaller, so their union is no larger than the bigger disc and nothing is gained. With c=1+2i the centres 1 and 1+2i are 2 apart, which is more than the difference of the radii, 22−2=0.8284, and less than their sum, 2+22=4.8284; so the two circles cross, neither disc contains the other, and D(1,2)∪D(1+2i,22) is strictly larger than either. That is a genuine enlargement of the domain.
They agree on the overlap. Both series sum to 1/(z+1) wherever they converge, so they agree on the lens D(1,2)∩D(1+2i,22), which is non-empty by the crossing just checked. Take the point z=0.2+1.6i: it lies in both, since ∣z−1∣=1.788854<2 and ∣z−1−2i∣=0.894427<2.8284. There
z+11=1.2+1.6i1=0.3−0.4i
exactly, since ∣1.2+1.6i∣2=4.
The convergence rates differ sharply. At that point the common ratios are −(z−1)/2=0.4−0.8i for f1, of modulus 0.894427, and −(z−c)/(c+1)=0.3−0.1i for f3, of modulus 0.316228. Counting the terms needed to bring the relative error below 10−7, f3 needs 14 and f1 needs 145; on an absolute criterion, error below 10−7 in modulus, the counts are 14 and 139. (We state the criterion because the two differ: the value has modulus 0.5, so a relative tolerance is the stricter one here.) With 200 terms each series returns 0.3000000−0.4000000i.
By part 3, f1 and f3 are analytic continuations of one another and together define 1/(z+1) on the union of the two discs. Repeating the construction with centres marching away from −1, always off the line through the previous centre and −1, continues the function to all of C∖{−1} — the chain of overlapping circles of Arfken, Weber and Harris (Arfken, Weber & Harris (2013), §11.6, pp. 504-505).
Example(The Gamma function: continuation by a functional relation)Original
Arfken, Weber and Harris (Arfken, Weber & Harris (2013), §11.6, pp. 506-507) point out that power series are not the only vehicle for continuation: a functional relation, an identity linking the values of a function at different arguments, does the job in one step. The Gamma function is the standard case.
The integral defines Γ only on a half-plane. For Rez>0 set
Γ(z)=∫0∞tz−1e−tdt.
Near t=0 the integrand has modulus tRez−1e−t, which is integrable exactly when Rez>0; the decay of e−t handles t→∞ for every z. So the integral converges absolutely on the open right half-plane H and diverges for Rez≤0.
Why it is analytic on H. Not by the Leibniz rule: differentiation under the integral sign is stated for a bounded interval and a real parameter, and its own note excludes both an unbounded interval and an unbounded integrand — and here both occur. Truncate and use Morera's theorem instead.
The truncations are entire. For n∈N put Γn(z)=∫1/nntz−1e−tdt. The integrand (z,t)↦e(z−1)lnte−t is continuous on C×[1/n,n] and the interval is compact, so Γn is continuous on C. Let γ be any closed contour in C. Writing the contour integral as an integral over a parameter interval turns ∮γΓn(z)dz into an iterated integral of a continuous function over a compact rectangle, where the order may be exchanged (Multiple integrals and change of variables):
∮γΓn(z)dz=∫1/nne−t(∮γtz−1dz)dt=0,
each inner integral vanishing by Cauchy's theorem, since for fixed t>0 the function z↦tz−1=e(z−1)lnt is entire. A continuous function all of whose closed-contour integrals vanish is analytic, by Morera's theorem (part 4 of the consequences in Cauchy's integral formula); so Γn is entire.
The truncations converge uniformly on closed substrips. Fix 0<a<b<∞ and let n≥1. For a≤Rez≤b we have ∣tz−1∣=tRez−1≤ta−1 when 0<t≤1 and ≤tb−1 when t≥1, so
Γ(z)−Γn(z)≤∫01/nta−1e−tdt+∫n∞tb−1e−tdt.
Both terms are tails of convergent integrals, so both tend to 0 as n→∞, and the bound does not depend on z. Hence Γn→Γ uniformly on {a≤Rez≤b}.
Morera again, for Γ itself. A uniform limit of continuous functions is continuous (Uniform convergence), and every point of H has such a strip around it, so Γ is continuous on H. If γ is a closed contour in H, its image is compact, hence contained in some strip a≤Rez≤b with 0<a<b; uniform convergence there lets the limit pass through the integral (Integrating and differentiating series term by term), so ∮γΓ(z)dz=limn∮γΓn(z)dz=0. By Morera, Γ is analytic on H.
Everything below — the functional relation, the uniqueness of the continuation, the poles and their residues — rests on this.
The functional relation. Integrate by parts on [ε,T] with u=tz, dv=e−tdt:
∫εTtze−tdt=[−tze−t]εT+z∫εTtz−1e−tdt.
For Rez>0 the bracket vanishes in the limits ε→0, T→∞, leaving
Γ(z+1)=zΓ(z),Rez>0.
Setting z=n and iterating from Γ(1)=∫0∞e−tdt=1 recovers Γ(n+1)=n!.
Continue. Rewrite the relation as Γ(z)=Γ(z+1)/z. The right-hand side makes sense whenever Rez>−1 and z=0, and it agrees with the integral on H, so by uniqueness (part 3) it is the analytic continuation of Γ to {Rez>−1}∖{0}. Iterating m times,
Γ(z)=z(z+1)⋯(z+m−1)Γ(z+m),
which continues Γ to {Rez>−m} minus the points 0,−1,…,−(m−1). Letting m→∞ gives a function analytic on C∖{0,−1,−2,…}, and the continuation is unique at every stage.
The poles and their residues. Near z=−k the factor (z+k) in the denominator is the only vanishing one, and the numerator Γ(z+k+1) is analytic and non-zero there, so Γ has a simple pole at −k. At a simple pole the factorisation f(z)=(z−z0)−1φ(z) with φ analytic and φ(z0)=0 (Classification of isolated singularities) identifies the Laurent coefficient a−1 as φ(z0)=limz→z0(z−z0)f(z), so
Both match the standard numerical values of the continued Γ to ten digits, at arguments where the defining integral diverges outright. Residues: at k=0 the predicted residue is 1, at k=1 it is −1, at k=2 it is 1/2.
Application(Dispersion relations, response functions and regularisation)Original
Dispersion relations. A causal response function is analytic in the upper half of the complex frequency plane. Its values on the real axis — what an experiment measures — therefore determine it throughout that half-plane, and the Kramers-Kronig relations, which reconstruct the real part of a susceptibility from the imaginary part, are a quantitative form of that determinacy. The identity theorem is what guarantees there is nothing left over to measure: one half of the response, known on the real axis, fixes the other.
Extending a convergent formula past its range. A partition function, a Feynman integral or a special function is frequently defined by an integral or series that converges only on part of the parameter domain, and then used outside it. Analytic continuation is the licence, and uniqueness is what makes the answer unambiguous. Dimensional regularisation is the textbook case: a loop integral is computed as a function of the space-time dimension d where it converges, expressed in Gamma functions, and continued in d; the poles of Γ located above are exactly the divergences that must be subtracted.
Resonances on the second sheet. Continuing a scattering amplitude from the physical region through a cut onto a second sheet reveals poles at complex energies — resonances with finite lifetimes. That the continuation is path-dependent (the monodromy theorem failing for non-homotopic paths around a branch point) is not a defect: the different sheets are different physics, and which sheet one is on is part of the answer.
Numerical practice. Continuing a function known only on a segment — by fitting a series and re-expanding about a shifted centre, as in the first example — is numerically unstable, because small errors on the segment are amplified. Uniqueness says the answer exists and is unique; it says nothing about conditioning, and this is why analytic continuation of noisy data (for instance from imaginary to real time in lattice simulations) is a genuinely hard inverse problem.
The integral of an analytic function over a closed contour equals 2πi times the sum of its residues at the enclosed singularities, each weighted by the winding number; at a pole of order n the residue is a single derivative of (z−z0)nf(z).
What it means. Cauchy's theorem says a closed contour integral vanishes when there is nothing singular inside. The residue theorem says what it equals when there is: one number per enclosed singularity, read off its Laurent series, times 2πi. Nothing about the contour matters except which singularities it winds around and how many times. This converts a large class of integrals — including real integrals that have no elementary antiderivative — into finite algebra.
Definition(Residue)AdaptedAdapted from Ahlfors (1979), Ch. 4, §5.1, Definition 3, p. 149
Let z0 be an isolated singularity of f, so that f is analytic on a punctured disc 0<∣z−z0∣<δ and has there a unique Laurent expansion f(z)=∑n∈Zan(z−z0)n (Laurent series). The residue of f at z0 is
z=z0Resf=a−1=2πi1∮∣ζ−z0∣=rf(ζ)dζ(0<r<δ).
Equivalently, and this is Ahlfors's formulation: the residue is the unique complex number R for which f(z)−R/(z−z0) is the derivative of a single-valued analytic function on the punctured disc. Here is why the two agree.
The number a−1 works. Write f(z)−a−1(z−z0)−1=∑n=−1an(z−z0)n and set
F(z)=n=−1∑n+1an(z−z0)n+1.
Both halves of this series converge on the punctured disc. For the non-negative powers this is because ∣n+1∣≥1 for every integer n=−1, so no coefficient is enlarged and the radius of convergence cannot shrink. For the negative powers, put w=(z−z0)−1: the principal part of f is ∑m≥1a−mwm, convergent for ∣w∣<1/R1, and the corresponding part of F is −∑m≥1(a−m−1/m)wm, whose coefficients are those of w−1∑k≥2a−kwk divided by m — again no enlargement, and the shift by one power changes nothing about where it converges. A power series may be differentiated term by term inside its disc of convergence (Power series), applied to each half separately. Hence F is analytic and single-valued on the punctured disc with F′=f−a−1(z−z0)−1.
No other number works. If f−R(z−z0)−1=G′ with G analytic and single-valued on the punctured disc, integrate over a circle ∣ζ−z0∣=r. The integral of a derivative of a single-valued analytic function over a closed contour is 0 (Contour integrals), so
0=∮f(ζ)dζ−R∮ζ−z0dζ=2πia−1−2πiR,
giving R=a−1.
Theorem(The residue theorem)AdaptedAdapted from Ahlfors (1979), Ch. 4, §5.1, Thm 17, p. 150, and the boundary form, p. 152
Let f be analytic on a region Ω except for isolated singularities at the points of a set S⊂Ω with no accumulation point in Ω. Let γ be a cycle lying in Ω∖S and homologous to zero with respect to Ω (The winding number). Then n(γ,a)=0 for only finitely many a∈S, and
2πi1∮γf(z)dz=a∈S∑n(γ,a)z=aResf.
Classical form. If γ is a positively oriented simple closed contour whose interior D together with γ lies in Ω, and f is analytic on D=D∪γ apart from finitely many isolated singularities a1,…,ak∈D, then every n(γ,aj)=1 and
Simple pole. If z0 is a pole of order 1,
z=z0Resf=z→z0lim(z−z0)f(z).
Pole of order n. If z0 is a pole of order at most n,
z=z0Resf=(n−1)!1z→z0limdzn−1dn−1[(z−z0)nf(z)].
Quotient with a simple zero in the denominator. If f=f1/f2 with f1,f2 analytic at z0, f1(z0)=0, f2(z0)=0 and f2′(z0)=0, then z0 is a simple pole and
z=z0Resf=f2′(z0)f1(z0).
Essential singularity. No general formula exists; the residue must be read off a Laurent expansion (Classification of isolated singularities). Arfken, Weber and Harris note (p. 510) that essential singularities do have well-defined residues, that in principle the coefficient integral can be used, and that the easiest route is often to find the Laurent expansion first.
Idea of the proofOriginal
Subtract off the bad part. Near each singularity a, the single term Ra/(z−a) with Ra the residue carries all of the obstruction to finding an antiderivative: by the definition above, f(z)−Ra/(z−a)is a derivative near a, so its integral over any small loop around a vanishes. Consequently the integral of f around a small loop at a is the integral of Ra/(z−a), which is 2πiRa.
The theorem is then pure bookkeeping on the contour. A cycle γ homologous to zero in Ω is, as far as integrals of functions analytic on Ω∖S are concerned, equivalent to the collection of small circles around the singularities, each counted n(γ,a) times — that is the content of the general Cauchy theorem. Summing the small-circle contributions gives the formula.
The computational rules come from the same place. Multiplying by (z−z0)n turns a Laurent series with a pole of order ≤n into a Taylor series in which the residue has moved to the coefficient of (z−z0)n−1, and Taylor coefficients are derivatives.
Step 1: only finitely many singularities count. The set U={a∈C∖γ:n(γ,a)=0} is open and contains everything outside a large circle (The winding number). Hence its complement K is closed and bounded, so compact. If infinitely many points of S lay in K they would accumulate somewhere in K; that accumulation point would be in Ω (because points of K with non-zero index cannot lie in the complement of Ω, γ being homologous to zero there), contradicting the hypothesis that S has no accumulation point in Ω. So n(γ,a)=0 for finitely many a∈S only; call them a1,…,ak and write nj=n(γ,aj).
Step 2: the local contribution of one singularity. For each j choose δj>0 so small that the punctured disc 0<∣z−aj∣<δj lies in Ω∖S, and let Cj be the circle ∣z−aj∣=δj/2, positively oriented. By the definition of the residue,
∮Cjf(z)dz=2πiz=ajResf.
Step 3: replace γ by the circles. Consider the cycle
σ=γ−j=1∑knjCj,
a cycle in the open set Ω′=Ω∖S. We claim σ is homologous to zero in Ω′, i.e. n(σ,b)=0 for every b∈/Ω′. There are two kinds of such b.
If b∈/Ω: then n(γ,b)=0 because γ∼0 in Ω, and n(Cj,b)=0 for each j because b lies outside the small disc bounded by Cj (that disc is contained in Ω). So n(σ,b)=0.
If b∈S: then n(Cj,b)=1 if b=aj and 0 otherwise, the discs being disjoint and small. Hence n(σ,b)=n(γ,b)−nj if b=aj for some j, which is 0 by the definition of nj; and n(σ,b)=n(γ,b)−0=0 for the remaining b∈S, since those have n(γ,b)=0 by Step 1.
Step 4: apply the general Cauchy theorem.f is analytic on Ω′ and σ∼0 in Ω′, so ∮σfdz=0 (The winding number). Therefore
which is the assertion, the omitted singularities contributing n(γ,a)=0. The classical form follows because a positively oriented simple closed contour has index 1 at every interior point and 0 outside (The winding number).
Step 5: the residue formulas. Let z0 be a pole of order at most n, so the Laurent series is f(z)=∑m≥−nam(z−z0)m. Then
is a convergent power series on the punctured disc; its sum extends analytically across z0, so it is the Taylor series of that extension (Taylor series of analytic functions). The coefficient of (z−z0)n−1 is a−1, and Taylor coefficients are derivatives:
a−1=(n−1)!1dzn−1dn−1[(z−z0)nf(z)]z=z0,
the derivative being that of the analytic extension, which is what the limit in the statement computes. Taking n=1 gives rule 1. For rule 3, write f2(z)=(z−z0)h(z) with h analytic and h(z0)=f2′(z0)=0 — possible since f2 has a simple zero (Taylor series of analytic functions). Then (z−z0)f(z)=f1(z)/h(z)→f1(z0)/f2′(z0), which is non-zero, so the pole is simple with that residue. □
Remark(Where the hypotheses are used)Original
"No accumulation point in Ω" is what Step 1 consumes. Its role is to make the residue sum finite. Step 1 shows that the points of S with non-zero index lie in a compact set K; if infinitely many points of S lay in K they would accumulate at a point of K, and that point would belong to Ω, which the hypothesis forbids. Without the hypothesis the sum on the right-hand side can have infinitely many non-zero terms and need not converge, so there is nothing left to prove or disprove. Note that the hypothesis does not exclude an accumulation of poles just outside Ω: for f(z)=1/sin(π/z) on Ω={0<∣z∣<2} the poles 1/n accumulate at 0, which is not in Ω, and the theorem applies there without difficulty to any cycle homologous to zero in Ω.
Homologous to zero, not merely closed. Take Ω={0<∣z∣<2}, f(z)=1/z2 (which has no singularity in Ω) and γ the circle ∣z∣=1. Then ∮γf=0, consistent with the theorem, but for f(z)=1/z we get 2πi although f has no singularity in Ω either. The resolution is that γ is not homologous to zero in this Ω: n(γ,0)=1 and 0∈/Ω. Dropping the homology hypothesis makes the theorem false.
Rule 2 needs only "order at most n". Using too large an n is harmless, merely more differentiation; using too small an n gives a wrong answer, because (z−z0)nf(z) then still has a pole and no Taylor series.
Rule 2 is often the hard way. Arfken, Weber and Harris warn on p. 510 that rule 2 can work out messier than a well-chosen series expansion, and they illustrate the point in item 4 of Example 11.7.1 on p. 511: for z/sin2z at z=π, differentiating is laborious, while substituting w=z−π and expanding (w+π)/sin2w gives the residue by inspection. Expanding is usually faster whenever the function is built from known series.
The theorem does not require poles. Essential singularities are included, and their residues are perfectly well defined; it is only the computation that has no closed formula.
Example(A contour integral with a double pole)Original
Evaluate
I=∮∣z∣=2z2(z−1)ezdz,
the circle positively oriented.
Locate the singularities. The integrand is analytic except where the denominator vanishes: z=0 (where z2 gives a pole of order 2, since ez/(z−1)→−1=0 there) and z=1 (a simple pole). Both lie inside ∣z∣=2, and neither lies on it. The contour is simple and positively oriented, so the classical form applies with both residues counted once.
As a cross-check by expansion: ez/(z−1)=(1+z+2z2+⋯)⋅(−1−z−z2−⋯)=−1−2z−⋯, so dividing by z2 gives −z−2−2z−1−⋯ and the residue is −2.
Assemble.
I=2πi(e−2)=2πi⋅0.7182818285=4.5130978310i.
Numerical check. Parametrising z=2eiθ and integrating numerically gives I=4.5130978310i to ten digits, with real part 10−22.
Example(Residues at a simple pole of a quotient, and at an essential singularity)Original
(a) Quotient rule. Find the residues of f(z)=z/(z4+4) at its poles. The poles are the fourth roots of −4, namely z=2eiπ/4,2e3iπ/4,2e5iπ/4,2e7iπ/4, i.e. ±1±i; each is simple because dzd(z4+4)=4z3=0 there. Rule 3 gives
z=zjResf=4zj3zj=4zj21.
At z1=1+i, z12=2i, so the residue is 1/(8i)=−i/8. At z2=−1+i, z22=−2i and the residue is i/8. The two poles in the upper half-plane thus have residues summing to 0 — which is why ∫−∞∞xdx/(x4+4) vanishes, as it must by oddness.
(b) Essential singularity. Find Resz=0z2sin(1/z). There is no formula to apply; expand. With sinζ=∑k≥0(−1)kζ2k+1/(2k+1)! at ζ=1/z,
z2sinz1=k=0∑∞(2k+1)!(−1)kz1−2k,
a Laurent series on 0<∣z∣<∞ with infinitely many negative powers, so z=0 is essential. The exponent 1−2k equals −1 only at k=1, so
z=0Resz2sinz1=3!(−1)1=−61.
Numerically, integrating z2sin(1/z) over ∣z∣=1 gives −1.0471975512i, and 2πi⋅(−1/6)=−1.0471975512i. Note the contrast with part (a): here the residue is not obtainable from any derivative formula, because no finite power of z removes the singularity.
Application(Where the residue theorem does the work)Original
Inverting transforms. The Bromwich inversion of a Laplace transform, f(t)=(2πi)−1∫Lf~(s)estds, and the inversion of the Z-transform are both evaluated by closing the contour and summing residues. Each pole of f~(s) contributes one exponential esjt with amplitude its residue: the poles are the normal modes, and the residues their amplitudes.
Mode sums and Green's functions. In a field-theory or electrodynamics calculation, a frequency or momentum integral of a propagator is done by residues, and the physical content — which modes propagate, with what weight, and with what causal structure — is read off from which poles the contour encloses. The choice of contour relative to the poles is exactly the choice of boundary condition (retarded, advanced, Feynman).
Counting and stability. Applied to f′/f, the theorem counts zeros and poles inside a contour (Argument principle and Rouche's theorem), which is the basis of the Nyquist stability criterion and of root-finding in a prescribed region.
An integral of a rational function of sine and cosine over one full period becomes a contour integral on the unit circle under z=eiθ, and is then 2πi times the sum of the residues, inside the circle, of the transformed integrand R(⋅,⋅)/(iz).
What it means. On the unit circle z=eiθ, the sine and cosine of θ are rational functions of z, and dθ is dz/(iz). So a real integral over one full period turns into a contour integral of a rational function around the unit circle — and a contour integral of a rational function is just a finite sum of residues. The method is completely mechanical: no antiderivative, no trigonometric identity, no partial fractions in θ. Its one real step is deciding which poles lie inside the circle.
Definition(The substitution)AdaptedAdapted from Arfken, Weber & Harris (2013), §11.8, Eqs. (11.92)-(11.93), pp. 522-523
Let R(s,c)=P(s,c)/Q(s,c) be a rational function of two variables such that θ↦R(sinθ,cosθ) is finite for every real θ, and consider
I=∫02πR(sinθ,cosθ)dθ.
Substituting z=eiθ, so that θ running from 0 to 2π traces the unit circle ∣z∣=1 once counterclockwise, and reading the result through the parametrisation definition of a contour integral (Contour integrals), which is what licenses the exchange of ∫02π⋯dθ for ∮⋯dz,
sinθ=2iz−z−1,cosθ=2z+z−1,dθ=izdz,
and more generally coskθ=21(zk+z−k), sinkθ=2i1(zk−z−k) for integer k. Hence
I=∮∣z∣=1F(z)dz,F(z)=iz1R(2iz−z−1,2z+z−1),
and F is a rational function of z.
The hypothesis on R is a condition along the real circle s2+c2=1, and it transfers directly to the unit circle in the z-plane: for ∣z∣=1 we have z=eiθ with θ real, so the two arguments of R in the display are the real numbers sinθ and cosθ, which satisfy s2+c2=1. Hence R is finite at every point of ∣z∣=1, and since the remaining factor 1/(iz) is finite there too, F has no pole on ∣z∣=1. That is the form in which the hypothesis is used below.
Method(Evaluating a trigonometric integral by residues)AdaptedAdapted from Arfken, Weber & Harris (2013), §11.8, Eqs. (11.92)-(11.94), pp. 522-523; Ahlfors (1979), Ch. 4, §5.3, pp. 155-156
Get a full period. If the integral is over an interval shorter than 2π, extend it using symmetry or periodicity before substituting. For instance, if the integrand is unchanged by θ↦2π−θ (true whenever it depends on θ only through cosθ), then ∫0π=21∫02π. If no such device works, this method does not apply directly.
Substitute. Replace sinkθ, coskθ and dθ by the expressions above and simplify to a single rational F(z). Keep coskθ as 21(zk+z−k) rather than first expanding it in powers of cosθ: it is less work.
Clear negative powers. Multiply numerator and denominator by the smallest power of z that makes F a ratio of polynomials. Do not forget the factor 1/(iz) from dθ: it is a genuine pole of the integrand at z=0 unless the rest of F cancels it. (Arfken states this as the rule that one needs the residues of f/z, not of f.)
Find the poles and keep the ones inside. Factor the denominator. Discard poles with ∣z∣>1. By the definition block there is no pole with ∣z∣=1, so every pole is strictly inside or strictly outside; if one turns up on the circle, the hypothesis that the denominator of R does not vanish on s2+c2=1 has failed, the original integral diverges, and at best a principal value can be assigned to it — which is a different method.
Sum the residues. By the residue theorem (The residue theorem),
I=2πi∣zj∣<1∑z=zjResF.
Check. The answer must be real whenever the original integrand is real. If R≥0 on the circle, the answer must be non-negative — and strictly positive unless R vanishes identically on the circle, since a continuous non-negative integrand with a non-zero value has a positive integral. A quick numerical quadrature of the original integral is the surest check.
Remark(Where the hypotheses are used, and useful short-cuts)Original
Finiteness of the integrand at every real θ is the one hypothesis to verify first: it is what keeps the poles of F off ∣z∣=1 and the original integral convergent. A convenient sufficient condition is that Q have no zero on the real circle s2+c2=1, but that is strictly stronger than needed, because a zero of Q there may be cancelled by a zero of P once s,c are restricted to the circle. For instance R(s,c)=c2/(1−s) has Q(1,0)=0, yet on the circle c2=(1−s)(1+s), so R(sinθ,cosθ)=1+sinθ is finite throughout and the integral is 2π. Riley, Hobson and Bence impose the condition directly in the z-plane, requiring the transformed integrand to have finitely many poles inside the circle and none on it (p. 861); with the hypothesis as stated here the two formulations agree.
Rationality in sinθ,cosθ is what makes F rational, hence single-valued with finitely many poles. An integrand such as 1+cosθ produces a branch point and is outside this method.
A full period is essential. Over a sub-interval the image of θ is an arc, not a closed curve, and there is no contour to apply the residue theorem to. Step 1 is therefore not a convenience but a requirement.
Reciprocal pairs of poles. When the denominator is a palindromic quadratic z2−2bz+1, its roots satisfy z1z2=1, so exactly one lies inside the unit circle unless both are on it. Noticing this saves computing both moduli. Arfken, Weber and Harris (Arfken, Weber & Harris (2013), Example 11.8.1, p. 523) make the step explicitly, writing that since z1z2=1 one root is within the unit circle and the other outside; Hassani (Hassani (2013), Examples 11.3.6-11.3.7, pp. 351-352) reaches the same conclusion in his parallel examples but states it directly, without invoking the product of the roots. All three examples below lean on the device — the third after the substitution u=z2.
Odd integrands integrate to zero. If R is odd under θ↦−θ, the integral over a full period vanishes, and the residue sum must come out zero; this is a good consistency check rather than a method.
Example(A full computation with a double pole at the origin)Original
Evaluate
I=∫02π2−cosθsin2θdθ.
Step 1. The range is already a full period. The denominator satisfies 2−cosθ≥1>0, so the integrand is finite everywhere and the integral converges.
Step 6: check.2π(2−3)=2π(0.2679492)=1.6835744, positive as it must be. Numerical quadrature of the original integral gives 1.6835744, and symbolic integration confirms the closed form 2π(2−3).
Example(A double pole away from the origin)Original
Evaluate
J=∫02π(2−cosθ)2dθ.
(This is a standard integral: Hassani (Hassani (2013), Example 11.3.7, pp. 351-352) works ∫0πdθ/(a+cosθ)2 for general a>1, of which the present J is twice the case a=2. The point here is the double pole away from the origin, and the steps and numbers below are our own.)
Substitute. Using 2−cosθ=−(z2−4z+1)/(2z) from the previous example,
with z±=2±3 as before. Note that the factor 1/z from dθ has been cancelled by the z2 from the squared denominator: there is no pole at the origin this time.
Check.43π/9=2.4183992. Numerical quadrature of the original integral returns 2.4183992, and symbolic integration gives 43π/9. The answer must exceed Jmin=2π/9=0.698 (the value if the denominator were always at its largest, 9) and be less than 2π=6.283 (its value if the denominator were always at its smallest, 1); it is.
Example(A quarter period, recovered by two symmetries)Original
Evaluate
K=∫0π/24−sin2θdθ.
Step 1: get a full period. The integrand depends on θ only through sin2θ, which has period π and is unchanged by θ↦π−θ. The reflection matches (0,π/2) with (π/2,π), so ∫0π=2∫0π/2; the periodicity then gives ∫02π=2∫0π. Hence
K=41∫02π4−sin2θdθ.
The denominator satisfies 4−sin2θ≥3>0, so the denominator of R does not vanish on the real circle and the integral converges.
Steps 2-3: substitute and clear. With z=eiθ, sin2θ=−41(z2−2+z−2), so
The factor 1/z from dθ has been cancelled, so there is no pole at the origin.
Step 4: poles. Put u=z2. Then u2+14u+1=0 gives u=−7±43, and the two values multiply to 49−48=1. Since (2−3)2=7−43, we get u=−(7−43) with square roots z=±i(2−3), of modulus 0.2679492<1, and u=−(7+43) with square roots of modulus 2+3=3.7320508>1. So exactly two poles lie inside the unit circle, namely z±=±i(2−3), both simple because z4+14z2+1 has four distinct roots. None lies on the circle.
Step 5: residues. The denominator has derivative 4z3+28z, so by the quotient rule (The residue theorem), at either inside pole zj,
z=zjResF=i(4zj3+28zj)4zj=i(zj2+7)1.
Both inside poles have zj2=−(7−43), hence zj2+7=43, and each residue is 1/(43i). The two together give 1/(23i).
Assemble.
K=41⋅2πi⋅23i1=43π=123π.
Step 6: check.3π/12=0.4534498; numerical quadrature of K returns 0.4534498, and of the full-period integral 1.8137994=4K. A crude bracket confirms the size: the denominator lies between 3 and 4, so K must lie between (π/2)/4=0.3927 and (π/2)/3=0.5236; it does.
Application(Orbital averages, lattice sums and filter design)Original
Averages over an orbit or a cycle. Time averages in celestial mechanics are integrals over one revolution of rational functions of the true anomaly: ⟨r−n⟩ with r=a(1−e2)/(1+ecosθ) is exactly of the form treated here, and the method delivers the classical results such as ⟨r−1⟩ and ⟨r−2⟩ in closed form in the eccentricity. The same integrals appear in the virial analysis of a Keplerian orbit and in radiation-reaction averages.
Angular averages in scattering and transport. The mean free path or the transport cross-section involves ∫02π or ∫0π of a rational function of cosθ — the Rutherford and screened-Coulomb kernels are of this type — and the residue method evaluates them without expanding in Legendre polynomials.
Lattice and Green's-function sums. The real-space Green's function of a tight-binding chain is G(n)=2π1∫02πE−2tcosθeinθdθ, which the substitution turns into a residue computation on the unit circle: the denominator becomes −t(z2−(E/t)z+1), a palindromic quadratic, so its two roots are a reciprocal pair. The dichotomy is therefore not inside versus outside — a reciprocal pair never has both roots outside, nor both inside — but on the circle versus off it. For ∣E∣>2∣t∣, outside the band, the roots are real, one strictly inside and one strictly outside, the hypothesis of the method holds, and G(n) decays geometrically in ∣n∣ like the inside root. For ∣E∣<2∣t∣, inside the band, the roots are e±ik with 2tcosk=E: both lie on the circle, the method does not apply as it stands, and the integral has to be defined by the retarded prescription E→E+i0, which pushes one root just inside and the other just outside and returns an oscillating, non-decaying G(n)∝eik∣n∣. So it is the band condition that decides where the reciprocal pair sits, and the i0 is not a decoration but the thing that makes the in-band Green's function well defined.
Digital filters. For a discrete-time filter with transfer function H(eiθ), the gain, the energy 2π1∫02π∣H∣2dθ and the autocorrelation coefficients are integrals of rational functions of eiθ over a period. The residue method is the standard evaluation, and the fact that the poles must lie strictly inside the unit circle is the stability condition for the filter.
A real integral over the whole line is computed by closing the contour with a large semicircle and summing the residues in that half-plane; the semicircle is negligible when zf(z)→0, and, for an integrand carrying a factor eiax, already when f(z)→0, which is Jordan's lemma.
What it means. The residue theorem needs a closed contour; a real integral has an open one. The fix is to add a large semicircular arc in the upper half-plane, apply the residue theorem to the closed contour, and then argue that the arc contributes nothing as its radius grows. How fast the integrand must decay for that argument to work is the whole content of the method. For a bare rational function one needs zf(z)→0. For an integrand with a factor eiaz, a>0, the exponential itself supplies decay in the upper half-plane, and f(z)→0 is already enough: that is Jordan's lemma, and it is what makes Fourier transforms of rational functions computable.
Definition(The contours and the arc)Original
For R>0 let ΓR denote the semicircular arc z=Reiθ, 0≤θ≤π, traversed from R to −R, and let CR be the closed contour consisting of the segment [−R,R] of the real axis followed by ΓR. Then CR is a positively oriented simple closed contour whose interior is the half-disc ∣z∣<R, Imz>0, so for a function f analytic on and inside CR apart from finitely many isolated singularities there,
∫−RRf(x)dx+∫ΓRf(z)dz=2πi∑Resf,
the sum over singularities in the open upper half-disc (Classification of isolated singularities). The lower-half-plane contour CR is built from the same two pieces, but the arc must now be traversed so that the two pieces join up: the real axis is run from −R to R, and then the arc ΓR runs from R to −R through −iR, returning to the starting point. Being closed is all that the residue theorem needs; what the orientation then decides is the sign. Walking from −R to R along the real axis, the enclosed lower half-disc lies on the right, so CR traversed in this order is negatively oriented — equivalently, the argument of z seen from an interior point decreases, from 0 at R through −π/2 at −iR to −π at −R, which is the clockwise sense. Hence
∫−RRf(x)dx+∫ΓRf(z)dz=−2πi∑Resf,
the sum now over singularities in the open lower half-disc. (To get a positively oriented boundary of the lower half-disc instead, run the real axis from R to −R and the arc from −R through −iR to R; that is the same identity with both sides negated.)
For an arc A⊆{∣z∣=R} write
MA(R)=z∈Amax∣f(z)∣,
and reserve the unsubscripted M(R)=max0≤θ≤π∣f(Reiθ)∣ for the upper semicircular arc ΓR, which is the arc Jordan's lemma is about. Statements made for a general arc below use MA; a lower-half-plane semicircle or a wedge of opening less than 2π gets its own maximum, which is in general unrelated to M(R).
Method(Vanishing of the arc for a rational-type integrand)AdaptedAdapted from Arfken, Weber & Harris (2013), §11.8, Eq. (11.96) and its proof, pp. 525-526
Let C=C(R) be the arc z=Reiθ with θ1≤θ≤θ2, the angles fixed, and let f be continuous on C(R) for every large R — so that ∣f∣ attains a finite maximum MC(R) on the compact arc, and the contour integral is defined at all. If limR→∞zf(z)=0 uniformly for z on those arcs — that is, RMC(R)→0 — then
R→∞lim∫Cf(z)dz=0.
For a rational f=p/q this holds exactly when degq≥degp+2.
Let a>0 and let f be analytic in the upper half-plane except at finitely many points, and continuous on each arc ΓR for all large R — which holds as soon as R exceeds the modulus of every one of those points, since f is then analytic, hence continuous, on ΓR, and ∣f∣ attains a finite maximum there. Assume
M(R)=0≤θ≤πmaxf(Reiθ)⟶0as R→∞.
Then
R→∞lim∫ΓReiazf(z)dz=0,
and quantitatively ∫ΓReiazf(z)dz≤πM(R)/a.
Only one half-plane. The hypothesis a>0 ties the conclusion to the upper half-plane: there ∣eiaz∣=e−aImz≤1, while in the lower half-plane the same factor grows. For a<0 the lemma applies instead to a semicircle in the lower half-plane.
Method(The resulting evaluations)AdaptedAdapted from Arfken, Weber & Harris (2013), §11.8, Eqs. (11.97) and (11.103), pp. 526, 529; Ahlfors (1979), Ch. 4, §5.3, cases 2 and 3, pp. 156-157
Rational integrand. Let f be analytic on R and in the upper half-plane except for finitely many isolated singularities, with RM(R)→0. Then the symmetric limit exists and
R→∞lim∫−RRf(x)dx=2πiImzj>0∑z=zjResf.
For a rational f=p/q with no real poles, the hypothesis amounts to degq≥degp+2; then ∣f(x)∣=O(x−2), so ∫−∞∞f converges absolutely and is the symmetric limit. Ahlfors notes (p. 156) that degq≥degp+2 is also exactly the condition for the real integral of a rational function to converge. Beyond the rational case, RM(R)→0 alone does not give convergence of ∫−∞∞f: see the remark block for an f, analytic on the closed upper half-plane with RM(R)→0, whose integral over [1,∞) diverges, so that only the symmetric limit exists.
Fourier-type integrand. Let a>0 and let f be analytic on R and in the upper half-plane except for finitely many isolated singularities, with M(R)→0. Then
∫−∞∞f(x)eiaxdx=2πiImzj>0∑z=zjRes(f(z)eiaz),
the integral being understood as limX1,X2→∞∫−X1X2, with the two limits taken independently. Taking real and imaginary parts gives ∫f(x)cosaxdx and ∫f(x)sinaxdx.
Idea of the proofOriginal
Both arc estimates are the ML inequality plus one extra idea.
For the rational case there is no extra idea: the arc has length πR and the integrand is at most M(R), so the integral is at most πRM(R), which is the hypothesis.
For Jordan's lemma that crude bound is useless, since RM(R) need not tend to 0. But ∣eiaz∣=e−aRsinθ is exponentially small except within O(1/(aR)) of the two ends of the arc, so the effective length of the arc is O(1/a), not πR. Making that precise needs only a lower bound on sinθ, and the elementary chord bound sinθ≥2θ/π on [0,π/2] does it: it turns the θ-integral into a geometric one whose value is (1−e−aR)π/(2aR), and the factor R cancels.
The second point, which Ahlfors emphasises, is about convergence rather than estimation: for a Fourier-type integrand with only a simple zero at infinity the integral converges only conditionally, so proving that the symmetric limit exists is not enough. Integrating over a rectangle instead of a semicircle lets the two ends go to infinity independently.
The last integrand is symmetric about θ=π/2, because sinθ is, so the integral equals twice the integral over [0,π/2]. On that interval sinθ is concave, hence lies above the chord joining (0,0) to (π/2,1):
sinθ≥π2θ,0≤θ≤2π.
(Concavity holds because (sinθ)′′=−sinθ≤0 there, and a concave function on an interval is at least its chord.) Therefore e−aRsinθ≤e−2aRθ/π and
Hence the arc integral is at most πM(R)/a, which tends to 0 with M(R). Note where a>0 entered: in the sign of the exponent −aRsinθ, which is non-positive only because sinθ≥0 on the upper arc and a>0.
Evaluation 1. Apply the residue theorem to CR for R large enough that every singularity of f in the upper half-plane lies inside (there are finitely many, so such R exists):
∫−RRf(x)dx+∫ΓRf(z)dz=2πiImzj>0∑z=zjResf.
The right-hand side is independent of R; the arc integral tends to 0 by the arc lemma applied to ΓR, for which MΓR(R)=M(R); hence ∫−RRf converges to the residue sum, which is the symmetric limit asserted. If moreover f=p/q is rational with degq≥degp+2, then ∣f(x)∣≤Cx−2 for large ∣x∣, so ∫−∞∞∣f∣<∞ and the symmetric limit is the integral. Nothing stronger is claimed in general, and nothing stronger is true: the remark block gives a non-rational f satisfying every hypothesis of part 1 for which ∫−∞∞f does not exist.
Evaluation 2. Jordan's lemma gives the same conclusion for the symmetric limit. To obtain the integral with independent limits, follow Ahlfors (p. 157) and replace the semicircle by three sides of a rectangle with vertices X2, X2+iY, −X1+iY, −X1, chosen large enough to contain all upper-half-plane singularities. The only hypothesis available is M(R)→0, so put
m(X)=R≥XsupM(R)=sup{∣f(z)∣:∣z∣≥X,Imz≥0},
which is finite for large X and tends to 0 as X→∞ precisely because M(R)→0. (Note m is non-increasing, and m(X)≥M(X).) Every point of the right-hand vertical side has ∣z∣≥X2, so ∣f∣≤m(X2) there, while ∣eiaz∣=e−ay; hence that side contributes at most
∫0Ym(X2)e−aydy<am(X2),
and similarly the left side contributes at most m(X1)/a. Every point of the top side has ∣z∣≥Y, so that side contributes at most m(Y)e−aY(X1+X2), which tends to 0 as Y→∞ for fixed X1,X2. Letting Y→∞ first and then X1,X2→∞ independently gives
Choose the half-plane. With a factor eiax present, the choice is forced: upper half-plane if a>0, lower if a<0. Without such a factor, either will do; pick the one with fewer or simpler poles.
Check the decay. Compute M(R). Use the arc lemma if RM(R)→0; otherwise use Jordan's lemma, which needs only M(R)→0 but requires the exponential factor.
Convert a cosine or sine to an exponential before closing the contour, not after: cosax and sinax both grow in both half-planes, so neither Jordan's lemma nor the arc lemma applies to them directly. Write ∫fcosaxdx=Re∫feiaxdx when f is real on the real axis, or symmetrise the range first.
Sum the residues in the chosen half-plane and multiply by 2πi (upper) or −2πi (lower).
Reduce a half-line integral to the whole line first, if possible. If the integrand is even, ∫0∞=21∫−∞∞. If it is neither even nor odd, this method may not apply; a sector contour may instead (Arfken, Example 11.8.6, p. 531).
Check. The answer must be real if the integrand is; it must have the right sign; and numerical quadrature settles it.
Remark(Where the hypotheses are used)Original
Decay, not merely convergence of the real integral.∫−∞∞sinx/xdx converges, but f(z)=sinz/z grows like e∣Imz∣ and no arc lemma applies. Rewriting it with eiz and a principal value is the correct route (Principal values and indented contours).
Jordan's lemma is strictly stronger where it applies. Riley, Hobson and Bence impose R times the maximum of ∣f∣ tending to zero as condition (ii) for closing with a semicircle (p. 862), and then make the comparison explicit when they come to Jordan's lemma (p. 864): there one needs only M(R)→0. The example ∫xsin2x/(x2+9)dx below has zf(z)→1=0, so the arc lemma fails and Jordan's lemma is required.
RM(R)→0 controls the arc, not the convergence of the real integral. Part 1 asserts only the symmetric limit, and outside the rational case that is all one gets. Take
f(z)=(z+i)Log(z+i)1,
with Log the principal branch. For Imz≥0 the point z+i has imaginary part at least 1, so it never meets the cut (−∞,0] and never equals 1: f is analytic on the whole closed upper half-plane, with no singularities at all. On ∣z∣=R in the upper half-plane, ∣z+i∣≥R−1 and ∣Log(z+i)∣≥ln(R−1), so RM(R)≤R/((R−1)ln(R−1))→0; numerically RM(R)=0.2171 at R=100 and 0.05429 at R=108, tracking 1/lnR. Every hypothesis of part 1 holds, and the symmetric limit is 0 because there are no residues — numerically ∫−RRf has modulus 0.225 at R=106 and falls like π/lnR, not like 1/lnR: the products ∣∫−RRf∣lnR come out as 3.0696 at R=104, 3.1086 at R=106 and 3.1228 at R=108, climbing towards π. (The constant matters only to keep the two decay rates quoted in this bullet distinct: the RM(R) values above are 1/lnR with constant 1.) But on the real axis f(x)∼1/(xlnx) as x→+∞, whose integral diverges: Re∫1Xf grows like lnlnX, taking the values 1.680 at X=102 and 3.289 at X=1010, a difference of 1.609 against lnln1010−lnln102=1.609. So ∫−∞∞f does not exist, while limR∫−RRf does. The two tails cancel; they do not separately converge.
Riley, Hobson and Bence corroborate the point from the other side: their recipe for closing with a large semicircle carries, besides the analyticity condition (i) and the decay condition (ii) quoted above, a third hypothesis (iii) that ∫−∞0f and ∫0∞fboth exist (p. 862). That is exactly what this example violates, and it is why our part 1 can only assert the symmetric limit: we have not assumed (iii), so we must not conclude anything stronger than what the contour gives.
a>0 and the choice of half-plane are not interchangeable. Closing ∫eixf(z)dz downwards makes ∣eiz∣=e+∣y∣ blow up, and the arc contribution diverges. Arfken states this as the remark that the two half-planes are not equivalent for Jordan's lemma (p. 528).
Finitely many singularities in the half-plane. Needed so that some finite R encloses them all. A function such as 1/sinh(πz) has infinitely many poles going up the imaginary axis, and this method must be replaced by one that controls an infinite residue sum.
No poles on the real axis. Otherwise the real integral diverges and only a principal value can be assigned.
Example(Rational integrand: a quartic denominator)Original
Evaluate
I=∫0∞x4+16x2dx.
Step 6 first: symmetrise. The integrand is even, so I=21∫−∞∞x2(x4+16)−1dx.
Steps 1-3. Put f(z)=z2/(z4+16). Its poles are the solutions of z4=−16, i.e. z=2ei(2k+1)π/4 for k=0,1,2,3:
all simple and none real. The upper half-plane contains z1 and z2. Since deg(denominator)=4≥2+2=deg(numerator)+2, the arc lemma applies; concretely ∣zf(z)∣∼∣z∣−1→0.
Step 5: residues. Each pole is a simple zero of z4+16, so by the quotient rule (The residue theorem)
Check.2π/8=0.5553603673; numerical quadrature of I returns 0.5553603673.
Example(Fourier-type integrand: Jordan's lemma doing work the arc lemma cannot)Original
Evaluate
J1=∫−∞∞x2+9cos2xdxandJ2=∫−∞∞x2+9xsin2xdx.
Set up both at once. Both are parts of ∫−∞∞g(x)e2ixdx with g(x) real: take g(z)=1/(z2+9) for J1 and g(z)=z/(z2+9) for J2, and use
J1=Re∫−∞∞x2+9e2ixdx,J2=Im∫−∞∞x2+9xe2ixdx,
valid because g is real on the real axis and e2ix=cos2x+isin2x.
Decay.z2+9 vanishes only at ±3i, so in both cases the only upper-half-plane singularity is the simple pole z=3i. For J1, M(R)=1/(R2−9)→0 and even RM(R)→0: either lemma works. For J2, M(R)=R/(R2−9)→0 but RM(R)→1=0: the arc lemma fails and Jordan's lemma is essential. Since a=2>0, we close upwards in both cases.
As a by-product, ∫−∞∞sin2x(x2+9)−1dx=0 and ∫−∞∞xcos2x(x2+9)−1dx=0, both correct by oddness.
Check. Oscillatory numerical quadrature returns 0.0025957432 and 0.0077872296. The exponential smallness is characteristic: the Fourier transform of a function analytic in a strip of width 3 about the real axis decays like e−3∣k∣, and here k=2 with the pole at distance 3 gives e−6.
Application(Fourier transforms, causal response and lattice Green's functions)Original
Fourier transforms of rational spectra. The pair 1/(x2+b2)↔(π/b)e−b∣k∣, computed above with b=3, is the Lorentzian-to-exponential pair that underlies linewidth analysis: a Lorentzian spectral line of half-width b corresponds to an exponentially decaying correlation function with rate b. The method generalises to any rational spectral density, which is how the correlation function of a linear system driven by white noise is computed.
Causality and which half-plane. A response function χ(ω) with no poles in the upper half-plane gives, by closing the contour upwards for t<0, the vanishing of the response before the stimulus. Jordan's lemma is the step that discards the arc, and the half-plane in which the exponential decays is fixed by the sign of t. The retarded, advanced and Feynman Green's functions differ only in how the contour passes the poles, and the different results are the different causal structures.
Lattice sums and propagators. Momentum integrals of the form ∫dkeikx/(k2+m2) produce the Yukawa factor e−m∣x∣ by exactly the computation above, with the pole at k=im fixing the range 1/m. The "mass equals inverse range" dictionary is a statement about the position of a pole in the complex momentum plane.
Deciding whether to use a sector instead. When the integrand is neither even nor odd and no exponential is available — for instance ∫0∞dx/(x3+1) — a wedge of opening 2π/3 replaces the semicircle and the same arc lemma applies to the arc of the wedge (Arfken, Example 11.8.6, p. 531). Recognising which contour closes an integral is the practical skill this method teaches.
For an integrand containing a fractional power or a logarithm, a keyhole contour hugging the cut converts the integral into a residue sum, because the two edges of the cut carry values of the multivalued factor differing by a known constant; a dogbone contour does the same for a cut joining two branch points.
What it means. A fractional power or a logarithm is not single-valued, so there is no contour integral to apply the residue theorem to until a branch has been chosen and a cut drawn. Far from being an obstacle, the cut is the mechanism. Run the contour along both sides of it: the integrand takes two different values on the two edges, differing by a factor e2πip for zp or by an additive 2πi for logz, so the two straight pieces do not cancel. What survives is a known multiple of the integral you wanted, and the residue theorem supplies its value.
Two shapes cover most cases: a keyhole, for a cut running from a branch point at the origin out to infinity, and a dogbone, for a cut joining two branch points, as in (z2−1)1/2.
Keyhole. Suppose the integrand has a single branch point at z=0 and the branch is fixed by taking argz∈(0,2π), so that the cut is the positive real axis. For 0<ρ<R the keyhole contourKρ,R consists of
the segment just above the cut, from ρ to R (where argz=0+);
the large circle ∣z∣=R, counterclockwise, from argz=0+ to argz=2π−;
the segment just below the cut, from R back to ρ (where argz=2π−);
the small circle ∣z∣=ρ, clockwise, closing the contour.
The two segments are a limit, and that is how they are to be read. The branch is undefined on the cut, so a contour with a piece lying on the positive real axis is not literally a contour in the domain of the integrand. What is meant is a limit. For small δ>0 let Kρ,Rδ be the honest closed contour made of the ray segment argz=δ, ρ≤∣z∣≤R, outwards; the circular arc ∣z∣=R from argz=δ to argz=2π−δ counterclockwise; the ray segment argz=2π−δ inwards; and the arc ∣z∣=ρ from 2π−δ back to δ clockwise. This lies entirely in the cut plane {0<argz<2π}, where the integrand is single-valued and analytic apart from its poles, so the residue theorem applies to it for each δ.
Letting δ→0+ is legitimate because the branch extends continuously to each edge of the cut. The integrand has finitely many poles and none on the cut, so their arguments avoid 0 and 2π: choose δ0>0 smaller than the argument of every pole and smaller than 2π minus it. Then (r,α)↦f(reiα) is continuous on each of the compact sets [ρ,R]×[0,δ0] and [ρ,R]×[2π−δ0,2π] once the two edge values of the jump statement below are assigned at α=0 and α=2π, hence uniformly continuous on each. (It need not be continuous on all of [ρ,R]×[0,2π], since a pole lies in between.) So the two ray integrals converge to the integrals of those edge values along [ρ,R], the two arc integrals converge to the arc integrals taken over the full angle, and the residue sum does not move at all, since no pole is crossed for small δ. We write Kρ,R for the limiting contour and use the edge values directly, which is what every computation below does.
Kρ,R is a positively oriented simple closed contour in the closed cut plane, and it encloses every singularity of the integrand with ρ<∣z∣<R other than those on the cut itself.
Dogbone. Suppose instead the multivalued factor is single-valued off a finite segment [α,β]⊂R — the standard case being (z2−1)1/2 with [α,β]=[−1,1]. The dogbone contourDε is the positively oriented boundary of an ε-neighbourhood of the segment: along the underside from α to β, round β, back along the top side from β to α, round α. It encloses the cut and nothing else.
Fix the branch of zp (p∈R) and of logz by argz∈(0,2π), that is, zp=∣z∣peipargz and logz=ln∣z∣+iargz. Fix x>0 and approach the point x of the cut along the circle of radius x, writing z=xeiα with α∈(0,2π). Then
(The two one-sided limits are the boundary values of the branch on the upper and lower edge of the cut; nothing depends on approaching along a circle rather than, say, vertically, since the branch is continuous up to each edge.)
Arfken stresses the rule that governs all such bookkeeping: on each part of the contour one must use the value of the multivalued factor reached continuously along the contour from the starting value, never a value that would require crossing the cut or an extra circuit of the branch point. For a pole at z0=∣z0∣eiθ0 with θ0∈(0,2π), the residue must therefore be computed with z0p=∣z0∣peipθ0 — for instance (−1)p read as eiπp and (−i)p as e3iπp/2, not e−iπp/2.
For (z2−1)1/2 with cut [−1,1] and the branch normalised by w(z)/z→1 as z→∞, the corresponding boundary values on the cut are
w(x+i0)=+i1−x2,w(x−i0)=−i1−x2,−1<x<1,
and w(x)=−x2−1 for real x<−1, w(x)=+x2−1 for real x>1.
Method(Convergence conditions)AdaptedAdapted from Ahlfors (1979), Ch. 4, §5.3, case 4, p. 159
Ahlfors treats ∫0∞xαR(x)dx for real α and rational R, observing that one may as well take α in the open interval (0,1). His stated requirements for convergence are two: at infinity R must vanish to order two or more, and at the origin R may have no singularity worse than a simple pole. He also records a different route to the same integral. Putting x=t2 rewrites it as 2∫0∞t2α+1R(t2)dt, and for the factor z2α one may then take the branch whose argument runs from −πα to 3πα, which is analytic once the negative imaginary axis is removed; the contour is then two segments along the real axis closed by a large and a small semicircle in the upper half-plane.
Name the branch and draw the cut so that the cut lies along the range of integration (Branch points and branch cuts). For ∫0∞ this means argz∈(0,2π) with the cut on the positive real axis.
Check the two circles vanish. Estimate ∣zf(z)∣ as ∣z∣→0 and as ∣z∣→∞; both must tend to 0, by the ML estimate (Contour integrals). A logarithm is harmless at the origin: on ∣z∣=ρ one has ∣logz∣≤∣lnρ∣+2π, so a factor logz against an otherwise bounded integrand gives ∣zf(z)∣=O(ρ∣lnρ∣)→0. (Hassani (Hassani (2013), §12.2, p. 371) states only that the circle contributions tend to zero, without the estimate; the estimate is ours.)
Write down both straight edges using the jump above. For a factor zp the lower edge returns −e2πip times the wanted integral I, so the two edges together give (1−e2πip)I. For a factor logz they give −2πiI~, where I~ is the integral of the rest of the integrand — the logarithms cancel and a new integral appears.
Sum the residues at all poles off the cut and multiply by 2πi (The residue theorem), computing every multivalued factor with arg∈(0,2π).
Solve the resulting linear equation for I.
Check.I must be real and of the right sign; numerical quadrature decides.
Arfken's two tricks. (i) If the integrand has a logarithm you do not want, note that the logarithms cancel between the edges and leave behind the integral of the rest: a logarithm can be inserted on purpose to make an otherwise cancelling pair of edges non-cancelling (Example 11.8.9, pp. 536-537). (ii) If the integrand has a logarithm you do want, use a contour that avoids the branch point and draw the cut outside the contour (Example 11.8.7, pp. 532-534).
Method(Dogbone recipe)Original
Normalise the branch of the two-branch-point factor by its behaviour at infinity, e.g. w(z)=(z2−1)1/2∼z, and record the boundary values on the cut (statement above).
Compute the two edges of the dogbone. With the values ±i1−x2 the two edges add rather than cancel, giving a known multiple of the wanted real integral. The small circles at α and β contribute nothing when the singularity there is weaker than ∣z−α∣−1, which an inverse square root is.
Blow the dogbone up to a large circle. The difference between the integral over ∣z∣=R and the integral over the dogbone is 2πi times the sum of residues at the poles lying between them (The residue theorem).
Estimate the large circle. If the integrand is O(∣z∣−2) its contribution tends to 0; if it is O(∣z∣−1) the circle contributes 2πi times the coefficient of 1/z in the expansion of the integrand at infinity, and that term must be kept. Beware the convention: in the usual one the residue at infinity is minus that coefficient, Resz=∞f=−a−1, so the circle contributes −2πiResz=∞f. We write the coefficient rather than the named residue, to keep the sign unambiguous.
Solve for the integral and check numerically.
Remark(Where the hypotheses are used)Original
The cut must lie along the range of integration (keyhole). Choosing argz∈(−π,π) instead puts the cut on the negative axis, the two edges of the contour then lie where there is nothing to integrate, and the method produces no information about ∫0∞.
p non-integral. If p is an integer, e2πip=1, the factor (1−e2πip) vanishes and the keyhole gives 0=0. The method needs the integrand to be genuinely multivalued; this is why Arfken's trick (i) inserts a logarithm when the natural integrand is single-valued.
Both circles must vanish, and this bounds p from both sides. For ∫0∞xp−1/(x+a)dx the small circle needs p>0 and the large circle needs p<1; outside 0<p<1 the real integral itself diverges at one end or the other.
Phases at the poles are the usual source of error. Arfken calls using e−iπ/2 in place of e3iπ/2 for (−i)p "a serious mistake" (p. 535), and his phase table for (z2−1)1/2 (p. 503) exists for the same reason: a pole's argument must be the one the contour actually reaches.
No pole on the cut. A pole on the cut is a pole on the contour and needs an indented contour and a principal value (Principal values and indented contours).
Encircling both branch points can restore single-valuedness. Arfken's Example 11.6.4 (p. 503) shows that a loop around both branch points of (z2−1)1/2 returns the original value, because each factor contributes a sign change. That is precisely why the dogbone, which encircles both, lives in a region where w is single-valued, and why the outward deformation in step 3 is legitimate.
Example(Keyhole with a fractional power)Original
Evaluate
I=∫0∞x+2x−1/3dx.
Step 1: branch. Take f(z)=z−1/3/(z+2) with argz∈(0,2π), so z−1/3=∣z∣−1/3e−iargz/3 and the cut is the positive real axis. Here p=−1/3 in the notation of the jump statement.
Step 2: circles. On ∣z∣=ρ, ∣zf(z)∣≤ρ⋅ρ−1/3/(2−ρ)∼ρ2/3/2→0. On ∣z∣=R, ∣zf(z)∣≤R⋅R−1/3/(R−2)∼R−1/3→0. Both circles drop out.
Step 3: edges. Above the cut z−1/3=x−1/3; below it z−1/3=x−1/3e−2πi/3. The upper edge runs from 0 to ∞ and the lower from ∞ to 0, so
The cancellation of e−iπ/3 is the signal that the phases were tracked consistently; a wrong phase at the pole leaves a complex answer.
Step 6: check.2−1/3=0.7937005 and sin(π/3)=0.8660254, so I=π⋅0.7937005/0.8660254=2.8792270. Numerical quadrature of the original integral returns 2.8792270. The general formula this reproduces is ∫0∞xp−1(x+a)−1dx=πap−1/sinpπ for 0<p<1: here p=2/3, a=2, and sin(2π/3)=sin(π/3).
Example(Keyhole with a square root and two poles)Original
Evaluate
J=∫0∞x2+9xdx.
Branch and circles. Take f(z)=z1/2/(z2+9) with argz∈(0,2π), so p=1/2. Small circle: ∣zf∣∼ρ3/2/9→0. Large circle: ∣zf∣∼R−1/2→0. The real integral converges for the same reasons: like x at 0 and like x−3/2 at infinity.
Edges. Below the cut z1/2=x1/2eiπ=−x1/2, so
∮Kfdz=J−eiπJ=2J.
(The edges add here, since e2πip=eiπ=−1.)
Residues. The poles are z=3i=3eiπ/2 and z=−3i=3e3iπ/2 — the second written with arg=3π/2, as the branch requires. Both are simple, and Res=z1/2/(2z)=1/(2z1/2), or directly z1/2/(2z):
Their sum is 63(e−iπ/4+e−3iπ/4)=63⋅(−i2)=6−i6, using e−iπ/4+e−3iπ/4=21−i+2−1−i=−i2.
Solve.
2J=2πi⋅6−i6=62π6,J=6π6.
Check.π6/6=1.2825498; numerical quadrature gives 1.2825498. Had the second pole been assigned arg=−π/2 instead of 3π/2, its residue would have had the phase e+iπ/4 and the two residues would have summed to 3⋅2/6, real, giving a purely imaginary J — the error Arfken warns about, immediately visible.
Example(Dogbone across a two-branch-point cut)Original
Evaluate
L=∫−11(x+2)1−x2dx.
Step 1: the branch. Let w(z)=(z2−1)1/2 with the branch cut [−1,1] and w(z)/z→1 as z→∞; concretely, w(z)=∣z−1∣1/2∣z+1∣1/2ei(θ1+θ2)/2 with θ1=arg(z−1) and θ2=arg(z+1) both taken in (0,2π). Where the formula is analytic to start with. The factor ∣z−1∣1/2eiθ1/2 is a branch of (z−1)1/2; it is analytic wherever θ1 is continuous, and the determination of arg with values in (0,2π) is continuous off the positive real axis of its own argument, i.e. off the ray [1,∞) in z. Likewise ∣z+1∣1/2eiθ2/2 is analytic off [−1,∞). Their product is w, so w is analytic on C∖[−1,∞) to begin with. The next paragraphs remove the two unwanted rays.
The ray (1,∞): continuity. There both arguments jump by 2π, so the half-sum jumps by 2π and ei(θ1+θ2)/2 is unchanged: w is continuous across (1,∞).
The ray (1,∞): analyticity, by Morera. Continuity is not yet analyticity, and steps 3-4 need analyticity, so finish the argument. Let x0>1 and let U be an open disc about x0 with U⊂{Rez>1}. On U the function w is continuous, by the previous paragraph, and analytic off the real diameter U∩R, by the paragraph before that. Let γ be a closed contour in U. Rather than perturb it, split it. Cut γ at its crossings of the diameter U∩R; each resulting arc lies in the closed upper half-disc or in the closed lower half-disc. Group the arcs accordingly and close each group by running along the diameter between consecutive crossings. Every added diameter segment is traversed once in each direction, once as part of the upper group and once as part of the lower, so the added contributions cancel in the sum and
∮γwdz=upper∑∮wdz+lower∑∮wdz.
Each closed contour on the right lies in a closed half-disc on which w is continuous and is analytic in the interior, so each of those integrals vanishes by Cauchy's theorem (Cauchy's theorem) applied on a slightly shrunken half-disc and passing to the limit, using the uniform continuity of w on the compact U. Hence ∮γwdz=0 for every closed contour in U, and a continuous function with that property is analytic, by Morera's theorem (part 4 of the consequences in Cauchy's integral formula). So w is analytic at x0, and the ray (1,∞) is not part of the cut.
The ray (−∞,−1). Nothing jumps there at all: the determination of arg in (0,2π) is discontinuous only on the positive real axis of its argument, and for real x<−1 both x−1 and x+1 are negative, so θ1 and θ2 are already continuous (both equal to π) in a neighbourhood. Thus w is analytic there by the first paragraph's reasoning applied to a disc about such a point, which meets neither [1,∞) nor [−1,∞).
The segment (−1,1). Only θ2 jumps, the half-sum jumps by π, and w changes sign. So the cut is exactly [−1,1].
Normalisation. Checking at z=iR: θ1,θ2→π/2, so w(iR)→Reiπ/2=iR, that is w(z)/z→1 along the positive imaginary axis. Step 4 needs this uniformly in direction, and one line upgrades it. Since w2=z2−1 identically, (w(z)/z)2=1−z−2, so for ∣z∣≥2
(w(z)/z)2−1=∣z∣21≤41,hencew(z)/z≥3/2.
Write u=w(z)/z. From the displayed bound, Re(u2)≥1−41>0 for ∣z∣≥2, so u2 lies in the open right half-plane, which forces argu into (−4π,4π) or into (43π,45π) — two disjoint sectors. Now u is continuous on {∣z∣≥2}, which is connected, since w is analytic off [−1,1]; and at z=2i the value of u is close to +1, hence in the first sector. A continuous function on a connected set cannot jump between two disjoint sectors, so argu∈(−4π,4π) throughout ∣z∣≥2. Combined with ∣u∣≥3/2 this gives Reu≥(3/2)cos(π/4)>21, so ∣u+1∣≥Re(u+1)>1 and
zw(z)−1=∣w/z+1∣(w/z)2−1≤∣z∣21(∣z∣≥2),
which is uniform in argz. In particular ∣w(z)∣≥23∣z∣ for ∣z∣≥2, which is the bound step 4 uses.
Boundary values on the cut. They are
w(x+i0)=i1−x2,w(x−i0)=−i1−x2(−1<x<1),
since there θ1=π, θ2=0+ above and θ2=2π− below; and at the real point z=−2, where θ1=θ2=π, we get w(−2)=3eiπ=−3.
Step 2: the two edges. Set f(z)=1/((z+2)w(z)) and let Dε be the dogbone around [−1,1], positively oriented: underside left to right, then topside right to left. Using 1/(−i)=i and 1/i=−i,
The two small arcs at ±1 contribute nothing: there ∣f∣=O(ε−1/2) while the arc length is O(ε), so their contribution is O(ε1/2).
Step 3: inflate to a large circle. The function f is analytic outside the cut except for the simple pole at z=−2. Between Dε and the circle ∣z∣=R (with R>2) lies exactly that pole, so
∮∣z∣=Rfdz−∮Dεfdz=2πiz=−2Resf.
Step 4: the large circle vanishes. By the uniform bound of Step 1, ∣w(z)∣≥23R at every point of ∣z∣=R≥4, while ∣z+2∣≥R−2. So ∣f∣≤2/(3R(R−2)) everywhere on the circle, and the ML estimate (Contour integrals) gives
∮∣z∣=Rfdz≤2πR⋅3R(R−2)2=3(R−2)4π⟶0.
Step 5: residue and answer.
z=−2Resf=w(−2)1=−31,
so 0−2iL=2πi(−1/3) and
L=3π.
Check.π/3=1.8137994; numerical quadrature of L gives 1.8137994. The sign of w(−2) was the only delicate point: taking w(−2)=+3 would give L=−π/3, negative, which is impossible for a positive integrand. More generally the same computation gives ∫−11dx/((x+c)1−x2)=π/c2−1 for c>1, which at c=2 is π/3.
Application(Spectral functions, aerofoils and dispersion integrals)Original
Band edges and square-root cuts. The density of states of a one-dimensional band, or the free propagator of a particle with a continuum threshold, carries a factor (E2−Δ2)−1/2 whose branch points are the band edges. Integrals of such quantities against a smooth weight are dogbone integrals. The weight in the formula computed above is (1−x2)−1/2, which normalised is the arcsine (Chebyshev) density μ(x)=1/(π1−x2), and that is exactly the density of states of the one-dimensional nearest-neighbour tight-binding chain, ρ(E)=1/(π4t2−E2), with half-bandwidth scaled to 1. So our result is the Stieltjes (Hilbert) transform of the one-dimensional band's density of states:
∫−11c−xμ(x)dx=c2−11(c>1),
which is 1/π times the integral computed above, since μ is even. This is the local Green's function of the chain outside the band, and the inverse square root at the band edge is the one-dimensional van Hove singularity.
It is worth being explicit about what this is not, because the two weights are easy to confuse. The Wigner semicircle is the different density ρsc(x)=π21−x2, with the square root in the numerator; its Stieltjes transform is
∫−11c−xρsc(x)dx=2(c−c2−1),
which at c=2 equals 0.535898, against 1/3=0.577350 for the arcsine law and π/3=1.813799 for the un-normalised integral above. No rescaling of the variable turns one weight into the other — (1−x2)−1/2 and (1−x2)+1/2 are different functions — so the semicircle law, the Bethe-lattice Green's function and the self-consistency condition of the coherent-potential approximation belong to the second formula, not the first. Both are dogbone integrals: put w(z)=(z2−1)1/2 in the denominator for the arcsine case and in the numerator for the semicircular one, and the same contour evaluates each.
Thin-aerofoil and crack problems. In two-dimensional potential flow and in linear fracture mechanics, the unknown is recovered from an integral equation whose kernel is 1/((x−ξ)1−ξ2) on the chord or the crack. Two qualifications matter, and both are about the hypotheses of this unit.
First, the inverse-square-root edge behaviour is the physically correct singularity at a crack tip and at an aerofoil's leading edge, but not at a sharp trailing edge: there the Kutta condition requires the vorticity to vanish, so the thin-aerofoil solution carries a factor (1−ξ)/(1+ξ), singular at the leading edge and zero at the trailing one, rather than the symmetric (1−ξ2)−1/2.
Second, the Glauert integrals of thin-aerofoil theory are principal values: the pole ξ=x sits on the cut, with x in the chord, which is precisely the case the dogbone recipe excludes (see the remark on poles on the cut) and which belongs to Principal values and indented contours. The dogbone computed above is the off-cut case c>1; the principal-value case is reached from it by letting c approach the cut from one side and taking the real part, which is how the Glauert results are recovered. The dogbone supplies the analytic function; the limit onto the cut supplies the aerofoil formula.
Sum rules with fractional powers. Integrals ∫0∞xp−1f(x)dx with non-integral p arise as Mellin transforms, in the theory of anomalous diffusion and in renormalisation-group matching. The keyhole evaluation is the standard tool, and the factor (1−e2πip)−1∝1/sin(πp) in the answer — with poles at integer p — is exactly the structure that reappears as the Γ-function poles of a Mellin transform.
Dispersion integrals. A response function with a branch cut along a continuum (rather than isolated poles) has its real and imaginary parts linked by an integral over that cut; deforming a contour onto the two edges of the cut is how such a relation is derived, and the jump across the cut, f(x+i0)−f(x−i0)=2iImf(x+i0), is precisely the discontinuity computed in the dogbone recipe.
A simple pole on the path of integration is handled by deleting a symmetric interval around it, which defines the Cauchy principal value; indenting the contour by a small semicircle contributes exactly half a residue, so the principal value equals 2πi times the residues in the half-plane plus iπ times those on the axis.
What it means. A simple pole sitting on the path of integration makes the integral divergent, but the divergence is one-sided: the integrand runs to +∞ on one side and to −∞ on the other. Deleting a symmetric interval around the pole and then shrinking it lets the two infinities cancel, and what is left is the Cauchy principal value — a finite number that the divergent integral does not have, but which is what physics almost always wants. In the complex plane the same symmetric deletion appears as a small semicircle, which contributes exactly half of what a full circle would. Hence the working rule: poles inside the contour count once, poles on it count a half.
Definition(Cauchy principal value)AdaptedAdapted from Arfken, Weber & Harris (2013), §11.7, Eqs. (11.69)-(11.71), pp. 512-513
Let f be integrable on every closed subinterval of [a,b] not containing the point x0∈(a,b). The Cauchy principal value of ∫abf is
when the limit exists. With several interior singularities, one δ is deleted symmetrically about each and the limit taken as all of them shrink. With an infinite limit of integration the deletion convention is extended in the same spirit: P∫−∞∞ means the symmetric limit limX→∞P∫−XX, so that a divergence at +∞ may cancel one at −∞ exactly as the two sides of a pole cancel. (Where the integral also converges with the two ends taken independently — as statement 2 below delivers for a Fourier-type integrand — the symmetric limit agrees with it; where it does not, the symmetric limit is what the contour method computes and the notation means.) It is also written p.v.∫ or with a bar through the integral sign.
It is not the integral. Arfken's example is ∫−abdx/x with a,b>0. The symmetric deletion gives (lnδ−lna)+(lnb−lnδ)=ln(b/a), independent of δ. But if the two cut-offs are allowed to shrink independently, δ1 on the left and δ2=2δ1 on the right, the same computation gives ln(b/a)−ln2: a different answer. The improper integral therefore has no value, and the notation presumes the location of the singularity is known.
Each is half, in magnitude, of the 2πia−1 that a full circuit would give, with the sign of the direction of travel; the two differ by exactly 2πia−1.
The indentation identity. Let f be analytic on a neighbourhood of the real axis except for a simple pole at x0, and let Cu be the real axis from −∞ to ∞ with the pole bypassed above by Cδ↑, Cd the same with the pole bypassed below. Then, when the principal value exists,
The two closed-contour calculations differ by 2πiResx0f, which is exactly the residue that one encloses and the other does not: Arfken's point (p. 515) that for the purpose of computing the principal value it makes no difference which way one passes the pole, even though the contour integrals themselves differ.
The evaluation rule. Let f be analytic in the closed upper half-plane except for finitely many isolated singularities with Imz>0 and finitely many simple poles on the real axis, and suppose the large semicircle contributes nothing (by the arc lemma, or by Jordan's lemma if a factor eiaz with a>0 is present; see Infinite real integrals and Jordan's lemma). Then
Ahlfors states this for f=R(z)eiz with R rational, and stresses two hypotheses: every real pole must be simple, and R(∞)=0. In words: a pole on the contour is counted with weight one half, as though half of it lay inside.
Method(Sokhotski-Plemelj formula)Original
Let x0∈R and let f:R→C be bounded and continuous, Lipschitz on a neighbourhood of x0, with ∫R∣f(x)∣/(1+∣x∣)dx<∞. Then
an identity used constantly in physics to separate the real (dispersive) from the imaginary (absorptive) part of a response.
Idea of the proofOriginal
Everything follows from one computation. On a small circle about a simple pole, the singular term a−1/(z−x0) contributes a−1iΔθ where Δθ is the angle swept — the modulus δ cancels between dz=iδeiθdθ and the denominator δeiθ — while the analytic remainder contributes O(δ) and dies. A half-circle sweeps π, hence half a residue; a full circle sweeps 2π. That single fact gives the indentation statement, and the evaluation rule is then the residue theorem applied to the indented closed contour.
The Sokhotski-Plemelj formula is the same half-residue seen in a different limit. Separating
x−x0∓iε1=(x−x0)2+ε2x−x0±(x−x0)2+ε2iε
displays the two terms of the answer: the first is an odd kernel that converges to the principal value because the oddness performs the symmetric cancellation automatically, and the second is a Lorentzian of unit area and shrinking width, i.e. an approximation to πδ.
The indentation. Write f(z)=a−1/(z−x0)+g(z) with g analytic, hence bounded by some M, on a disc about x0. Parametrise Cδ↑ by z=x0+δeiθ with θ running from π down to 0; then dz=iδeiθdθ and
exactly, for every δ. The remainder obeys ∫Cδ↑g≤πδM→0 by the ML estimate (Contour integrals). For Cδ↓ the angle runs from π up to 2π, giving +iπa−1. Note what made the modulus drop out: the pole is simple.
What happens at a higher-order pole. The same parametrisation applied to a single term a−m(z−x0)−m with m≥2 gives, exactly,
The constant (1−(−1)1−m)/(1−m)vanishes for every odd m and equals −2/(m−1) for every even m. So the odd terms of the principal part contribute nothing at all to the arc, however high their order, while each even term contributes −2a−mδ1−m/(m−1), which diverges as δ→0+. (Checked symbolically for m≤7: the arc integral of (z−x0)−m is 0 for m=3,5,7 and −2/δ, −2/(3δ3), −2/(5δ5) for m=2,4,6.) Hence the indentation has a finite limit exactly when a−2=a−4=⋯=0, and that limit is still −iπa−1. Simplicity of the pole is the easy sufficient condition, not the exact one; the recipe's step 1 returns to this.
Part 1 of the indentation statement.Cu consists of (−∞,x0−δ], the arc Cδ↑, and [x0+δ,∞). Its integral is therefore the sum of the two straight pieces and the arc; letting δ→0+, the straight pieces converge to the principal value by definition and the arc to −iπa−1. Same for Cd with +iπa−1.
Part 2, the evaluation rule. Close Cu with the large semicircle ΓR of Infinite real integrals and Jordan's lemma. For R large and δ small, the resulting closed contour is positively oriented and encloses exactly the singularities with Imz>0: the real pole has been excluded by indenting above it. By the residue theorem (The residue theorem),
∫Cu(R,δ)fdz+∫ΓRfdz=2πiImzj>0∑z=zjResf.
Letting R→∞ kills the arc by hypothesis, and letting δ→0+ turns the first term into P∫f−iπResx0f by part 1. Rearranging gives the rule, extended to several real poles by indenting at each. Indenting below instead encloses the real pole as well, adding 2πiResx0f to the right-hand side and +iπResx0f to the left; the two statements are equivalent, since 2πia−1−iπa−1=iπa−1. □
Proof(Sokhotski-Plemelj)Original
Shift so that x0=0 and write g(u)=f(x0+u). Rationalising the denominator,
The integrand is dominated by ∥g∥∞/(1+t2), which is integrable, and converges pointwise to g(0)/(1+t2) by continuity of g at 0. By dominated convergence the limit is g(0)∫dt/(1+t2)=πg(0)=πf(x0).
The Kε term gives the principal value. Fix δ>0 small enough that g is Lipschitz with constant L on [−δ,δ]. Split the integral at ∣u∣=δ.
The first term is at most ε2δ−3⋅2ε−1∥g∥∞=2ε∥g∥∞/δ3→0; the second tends to 0 because ∫∣g(u)∣/(1+∣u∣)du<∞ makes the tail integrable. So the outside part converges to ∫∣u∣≥δg(u)du/u.
Inside. Both Kε and 1/u are odd, so ∫−δδKε(u)du=0 and the principal value of ∫−δδdu/u is 0. Subtracting the constant g(0) therefore costs nothing:
∫−δδKε(u)g(u)du=∫−δδKε(u)(g(u)−g(0))du,
and since ∣g(u)−g(0)∣≤L∣u∣ and ∣Kε(u)∣≤1/∣u∣, the integrand is bounded by L and the whole inside part is at most 2Lδ in modulus, uniformly in ε. The same bound holds for P∫−δδg(u)du/u.
Combine. For every δ,
ε→0+limsup∫Kεg−P∫ug(u)du≤4Lδ,
and δ was arbitrary, so the limit is the principal value. Adding ±iπf(x0) from the other term completes the proof. □
Confirm the real pole is simple. That is the condition under which everything below is stated. If the order is ≥2, the exact condition for both the principal value and the indentation to exist is that the even-index coefficients a−2,a−4,… of the Laurent expansion at the real pole all vanish, as the proof shows; an odd-order pole with no even-index terms is therefore still tractable, and the rule of step 5 applies to it unchanged. Arfken, Weber and Harris set this case as Exercise 11.7.11 (p. 522).
Decide what you are computing. If the integral diverges, say so and state that you are computing its principal value. For ∫fcosax or ∫fsinax with a pole at a zero of the trigonometric factor, the integral may converge outright and no principal value is needed.
Indent above each real pole and close with a large semicircle in the upper half-plane. Verify that the large arc vanishes.
Apply the rule:P∫=2πi∑(residues with Imz>0)+iπ∑(residues on the axis).
Take the real or imaginary part as required, and check reality and sign. For a numerical check, evaluate T(δ)=∫−∞x0−δ+∫x0+δ∞ at a small δ. Writing f=a−1/(x−x0)+g with g regular at x0, the deleted piece contributes nothing from the singular term (it is odd) and 2δg(x0)+O(δ3) from g, so
T(δ)=P∫f−2δg(x0)+O(δ3),
a prediction one can compare with the observed discrepancy.
Remark(Where the hypotheses are used)Original
Symmetric deletion is the definition, and it matters. Arfken's ∫−abdx/x shows that letting the two cut-offs shrink at different rates changes the answer by ln(δ2/δ1). The principal value is a choice, justified in applications by the physics (a symmetric regularisation of a resonant denominator) and not by the integral itself.
Simplicity of the real pole is sufficient in both places, and in neither is it the exact condition. In the arc computation simplicity is what makes the two powers of δ cancel and leaves the clean −iπa−1. In the definition it is what makes the two one-sided divergences logarithmic and of opposite sign. But the precise condition is the same in both: the symmetric deletion of ∫a−m(x−x0)−mdx gives 0 for odd m, the integrand being odd, and diverges like 2a−mδ1−m/(m−1) for even m≥2 — matching term by term the arc computation in the proof. So a real pole of order 3 with a−2=0, say f(x)=(x−x0)−3+g(x) with g analytic, has a perfectly good principal value and obeys the rule of step 5 with the same iπa−1; a pole of order 2 does not. Simplicity is the condition worth checking in practice because it is immediate, not because it is necessary.
Above or below: the principal value does not care. The two contour integrals differ by a full residue, but so do the two residue sums, and the principal value comes out the same. This is Arfken's remark after Eqs. (11.77)-(11.78).
The ±iε prescription is a choice of physics, not of mathematics. Sokhotski-Plemelj shows that the two prescriptions differ by 2πif(x0), a genuine difference: retarded versus advanced response, outgoing versus incoming waves, decaying versus growing modes. Which sign to use is fixed by causality or by boundary conditions, never by the integral.
Lipschitz at x0, not merely continuous. Without some modulus-of-continuity control at x0 the principal value itself may fail to exist; the Lipschitz bound is what makes the inside contribution O(δ) uniformly in ε.
Example(A rational principal value)Original
Evaluate
I=P∫−∞∞(x2+4)(x−1)xdx.
Step 1-2. Put f(z)=z/((z2+4)(z−1)). Singularities: z=±2i (simple, off the axis) and z=1 (simple, on the axis, with 1/(1+4)=1/5=0, so a genuine pole). The integral diverges; we compute its principal value.
The imaginary parts cancel, as they must for a real integrand — a built-in check that the half-residue was included with the right sign.
Step 6: check.2π/5=1.2566371. Deleting (1−δ,1+δ) with δ=10−3 and integrating numerically gives T(δ)=1.2563971. To predict the discrepancy, split f(x)=h(x)/(x−1) with h(x)=x/(x2+4); the regular part of f at x=1 is (h(x)−h(1))/(x−1), whose value at x=1 is h′(1). Since h′(x)=(4−x2)/(x2+4)2, we get h′(1)=3/25=0.12, so step 6 of the recipe predicts T(δ)=2π/5−2(0.001)(0.12)=1.2566371−0.00024=1.2563971 — matching every digit computed.
Example(A Fourier-type principal value, and where the half-residue is the whole answer)Original
Evaluate
J=P∫−∞∞x−2sin3xdx.
Set up. Take f(z)=e3iz/(z−2), whose only singularity is the simple pole at z=2 on the axis. Here M(R)=1/(R−2)→0, so Jordan's lemma applies with a=3>0 (Infinite real integrals and Jordan's lemma) and the large upper semicircle vanishes. There are no singularities in the open upper half-plane at all, so
P∫−∞∞x−2e3ixdx=0+iπz=2Resz−2e3iz=iπe6i.
The entire answer is the half-residue from the pole on the contour.
the first integral being the Dirichlet integral (which converges without a principal value, the singularity being removable) and the second vanishing by oddness. Numerically ∫−∞∞sin3udu/u=3.1415927=π and πcos6=3.0164639, so J=3.0164639. Note that J itself converges slowly under naive truncation, because the integrand decays only like 1/x; the substitution route is the reliable numerical check.
Example(The ±iε prescription in action)Original
Take f(x)=1/(1+x2) and x0=0, and compare the two sides of the Sokhotski-Plemelj formula.
Left-hand side in closed form. For 0<ε<1 let
A(ε)=∫−∞∞(1+x2)(x−iε)dx.
The integrand is O(∣z∣−3), so the large semicircle vanishes and the upper half-plane encloses the two simple poles z=i and z=iε. Their residues are
Right-hand side. The principal value P∫dx/((1+x2)x) vanishes, the integrand being odd; and iπf(0)=iπ. So the formula predicts A(ε)→iπ, which the closed form iπ/(1+ε) confirms, with the approach linear in ε.
Numerically, direct quadrature of A(ε) gives 2.8559933i at ε=0.1, 3.1104878i at ε=0.01 and 3.1384542i at ε=0.001, against iπ=3.1415927i; the closed form π/(1+ε) gives 2.8559933, 3.1104878 and 3.1384542 — agreeing to all digits shown. Choosing x−iε→x+iε instead flips the sign to −iπ, a difference of 2πif(0), exactly as the formula says.
Application(Dispersion relations, propagators and resonant denominators)Original
Kramers-Kronig relations. For a causal response χ(ω), Cauchy's formula applied to a contour along the real axis indented at ω gives
Reχ(ω)=π1P∫−∞∞ω′−ωImχ(ω′)dω′,
and the half-residue is exactly the term that separates the two parts. Every dispersion relation in optics, acoustics and particle physics has a principal value in it for this reason.
Propagators and the iε prescription. The retarded Green's function of the wave equation, the Feynman propagator and the advanced propagator differ only in how the contour passes the poles at ω=±k, that is, in the sign of iε in 1/(ω2−k2±iε). Sokhotski-Plemelj translates the prescription into a principal value plus a delta function, and the delta function is what carries the on-shell, energy-conserving part of the physics.
Fermi's golden rule. Second-order perturbation theory produces 1/(Ei−Ef+iε); applying Sokhotski-Plemelj splits it into a principal value, which renormalises the energy (the level shift), and −iπδ(Ei−Ef), whose modulus squared gives the transition rate. The golden rule is the imaginary half of this identity.
Landau damping and plasma dispersion. The dielectric function of a warm plasma contains ∫dv(∂f0/∂v)/(ω−kv), whose pole at the resonant velocity sits on the integration path. Taking the principal value plus the half-residue gives the real frequency shift plus the Landau damping rate; the sign of the iε — fixed by the initial-value nature of the problem — is what makes the damping a damping rather than a growth.
Multiplying a function by πcotπz, whose residue at every integer is 1, turns an infinite sum over the integers into a contour integral that vanishes; the sum is then minus the residues at the poles of the function, and πcscπz does the same for alternating sums.
What it means. A contour integral that vanishes is a statement that a sum of residues is zero. If the integrand has one simple pole at each integer, with residue f(n), and a few more poles elsewhere, then that statement reads: the whole infinite sum ∑nf(n) equals minus the finitely many remaining residues. The kernel that supplies a residue 1 at every integer is πcotπz; replacing it by πcscπz gives residue (−1)n instead and therefore sums alternating series. The work is then a handful of residue computations, and the answers come out in closed form — coth and csch for the two kernels — where elementary methods give nothing.
For N∈N0 let SN be the positively oriented boundary of the square with vertices ±(N+21)±i(N+21). It encloses exactly the integers −N,…,N and passes through no integer.
Method(The kernels are bounded on the squares)Original
On SN, for every N≥0,
∣κ(z)∣≤πcoth2π≈3.4254,∣κalt(z)∣≤π.
In particular both are bounded on ⋃NSN by a constant independent of N.
the sums on the right running over the poles zk of f.
If f has a pole at an integer m. The formula still holds in the form "all residues of the integrand sum to zero", with the integer m now contributing the residue of κf at m rather than the term f(m): that is,
n∈Zn not a pole of f∑f(n)+k∑z=zkRes[κ(z)f(z)]=0,
where now the zk include the integer poles of f. The same holds with κalt.
Idea of the proofOriginal
Build a function whose residues are the terms of the series. Since sinπz has a simple zero at each integer, π/sinπz and πcotπz have a simple pole at each integer, and their residues are (−1)n and 1. Multiplying by f, which is regular at the integers, gives an integrand with residue (−1)nf(n) or f(n) there — the terms of the series, laid out one per pole.
Now integrate over a contour that grows to swallow the whole integer lattice. The residue theorem equates the integral to 2πi times (the partial sum plus the finitely many residues at the poles of f). If the integral tends to zero, the infinite sum must equal minus those finitely many residues.
For the integral to tend to zero, two things are needed: the kernel must stay bounded on the growing contour — which it does, because cot and csc approach ∓i and 0 as ∣Imz∣ grows, and are controlled on the vertical sides by the half-integer offset — and f must decay faster than 1/∣z∣, so that the perimeter growth is beaten.
Step 1: residues of the kernels. The function sinπz has a zero at each integer n, and dzdsinπzz=n=πcosπn=π(−1)n=0, so the zero is simple and these are all its zeros. By the quotient rule for residues (The residue theorem),
(Riley, Hobson and Bence obtain the first by writing z=n+ξ and expanding, which gives cotπz≈1/(πξ); the two routes agree.) Since cosπz and sinπz have no common zero, the poles of both kernels are exactly the integers.
Step 2: the bounds on SN. Write z=x+iy. From sinπz=sinπxcoshπy+icosπxsinhπy and cosπz=cosπxcoshπy−isinπxsinhπy,
∣sinπz∣2=sin2πx+sinh2πy,∣cosπz∣2=cos2πx+sinh2πy,
using cosh2=1+sinh2. Hence
∣cotπz∣2=sin2πx+sinh2πycos2πx+sinh2πy.
Horizontal sides, ∣y∣=N+21≥21: dropping sin2πx≥0 below and bounding cos2πx≤1 above,
∣cotπz∣2≤sinh2πy1+sinh2πy=coth2π∣y∣≤coth22π,
coth being decreasing on (0,∞). Likewise ∣cscπz∣≤1/∣sinhπy∣≤1/sinh(π/2)<1.
Vertical sides, x=±(N+21): then sin2πx=1 and cos2πx=0, so
Since coth(π/2)=1.0903>1, the stated bounds hold on all of SN: ∣κ∣≤πcoth(π/2) and ∣κalt∣≤π. The half-integer offset is what forces sin2πx=1 on the vertical sides and so keeps the contour away from the zeros of sinπz; a square through the integers would give no bound at all.
Step 3: the integral vanishes. Let g=κf and take N large enough that all poles of f lie inside SN. The perimeter of SN has length 8(N+21), and every point of SN has ∣z∣≥N+21. By the ML estimate (Contour integrals),
where the last step uses (N+21)∣f(z)∣≤∣zf(z)∣ on SN. By hypothesis the right-hand side tends to 0 as N→∞. The same bound holds with κalt and the constant π.
Step 4: apply the residue theorem.SN is a positively oriented simple closed contour, and the integrand κf is analytic on and inside it except at the integers −N,…,N — where, f being analytic there, the residue is f(n)⋅1=f(n) — and at the poles zk of f. Hence (The residue theorem)
Letting N→∞, the left-hand side tends to 0 by Step 3, so the symmetric partial sums converge and
n=−∞∑∞f(n)=−k∑z=zkRes[κf].
Step 5: the two series converge absolutely. So far only the symmetric partial sums have been shown to converge. In fact the stated hypotheses force f to be a rational function, and then absolute convergence is immediate. Let z1,…,zJ be the poles of f, with principal parts Pj at zj — each Pj a finite sum of negative powers of z−zj (Classification of isolated singularities) — and put
P=j=1∑JPj,h=f−P.
Each Pj is analytic off zj, and f−Pj has a removable singularity at zj by the definition of the principal part, so h extends to an entire function. Both f and P tend to 0 as ∣z∣→∞ — f because zf(z)→0, and P because every term is a negative power — so h(z)→0 at infinity and is therefore bounded on C. By Liouville's theorem (Liouville's theorem and the maximum modulus principle) h is constant, and the constant is its limit at infinity, namely 0. Hence f=P is rational.
Write f=p/q in lowest terms. Then zf(z)→0 forces degq≥degp+2, so there are C and N0 with ∣f(x)∣≤Cx−2 for real ∣x∣≥N0. Therefore ∑n∣f(n)∣<∞: both series converge absolutely, and the symmetric limit found in Step 4 is their sum. The alternating case is identical with κalt, whose residue at n is (−1)n, and ∣(−1)nf(n)∣=∣f(n)∣.
The case of a pole at an integer. Nothing in Step 4 used that the integers are not poles of f except the identification of the residue at n as f(n). If f has a pole at m∈Z then κf has a pole of higher order there, and that point simply joins the list of zk; the displayed identity becomes the statement that all residues of κf sum to zero. □
Write the series as ∑n∈Zf(n) or ∑n∈Z(−1)nf(n) for a single meromorphic f. A one-sided sum ∑n≥1 is obtained afterwards from the two-sided one using the symmetry of f; for even f, ∑n≥1f(n)=21(∑n∈Zf(n)−f(0)).
Check the decay. Verify zf(z)→0; for a rational f=p/q this means degq≥degp+2.
Choose the kernel:πcotπz for a plain sum, πcscπz for an alternating one.
Find the poles of f and compute the residue of (kernel)×f at each, using cot(iu)=−icothu and csc(iu)=−icschu when the poles are imaginary. If f has a pole at an integer, include it and remember to omit that term from the sum.
The two-sided sum is minus the total of those residues.
Check. The answer must be real and of the right sign; summing twenty or so terms numerically settles it. A useful limiting check: as the parameter that moves f's poles to infinity is taken to its limit, the closed form should reduce to something known.
Remark(Where the hypotheses are used)Original
Half-integer sides. The whole method rests on the contour avoiding the zeros of sinπz by a fixed margin. On a square through the integers the kernel is unbounded and Step 3 fails.
zf(z)→0, not merely f→0. The perimeter grows like N, so f must beat 1/N. For f(z)=1/z the method fails — as it must, since ∑n=01/n is only conditionally summable and ∑n∈Z is not defined.
The hypotheses are more restrictive than they look. Step 5 shows that a function meromorphic on C with finitely many poles and zf(z)→0 is necessarily rational, with degq≥degp+2. So the natural domain of the method is rational summands, and the convergence of the two series is not an extra assumption but a consequence. A summand like 1/(en+1), decaying far faster, is not covered: it is not the restriction to Z of a function with finitely many poles, and summing it needs a kernel adapted to its own poles.
No pole of f at an integer, unless handled explicitly. Otherwise a term of the series is infinite. The corrected form of the statement is what makes the classical evaluation of ∑1/n2 work, where f=1/z2 has its pole exactly at n=0.
Boundedness of the kernel is uniform, the decay of f is not automatic. Riley, Hobson and Bence use a large circle rather than a square and argue that cotπz→∓i as Imz→±∞ (p. 883). On a circle one must still keep away from the real axis near the integers; the square makes that control explicit, which is why we use it.
csc rather than cot for alternating sums because its residue carries the sign: Riley, Hobson and Bence record exactly this ((−1)nπ−1 for cscπz, p. 883).
Here zf(z)=z/(z2+4)→0, and f has simple poles at z=±2i, neither an integer. Both hypotheses hold.
Steps 3-4: residues. With κ(z)=πcotπz and z2+4=(z−2i)(z+2i),
z=2iRes[κf]=2⋅2iπcot(2πi)=4iπcot(2πi).
Now cot(iu)=cos(iu)/sin(iu)=coshu/(isinhu)=−icothu, so cot(2πi)=−icoth2π and
z=2iRes[κf]=4iπ(−icoth2π)=−4πcoth2π.
At z=−2i: cot(−2πi)=+icoth2π (the cotangent is odd) and the denominator is −4i, giving the same value −41πcoth2π. The two residues are equal, as the evenness of f and the oddness of cot together require.
Step 5.
n∈Z∑n2+41=−(−4πcoth2π−4πcoth2π)=2πcoth2π,
and therefore
Σ1=21(2πcoth2π−41)=4πcoth2π−81.
Step 6: check.coth2π=1.0000069747, so Σ1=0.7854036413−0.125=0.6604036413. Direct numerical summation gives 0.6604036413. Two sanity checks: coth2π is barely above 1, so Σ1≈π/4−1/8=0.6603982, and the first four terms already give 1/5+1/8+1/13+1/20=0.4519231, below the total as they must be. The general formula is ∑n∈Z(n2+a2)−1=(π/a)cothπa; the limit a→0 reproduces the divergence of ∑n−2+a−2 correctly, since (π/a)cothπa=a−2+π2/3+O(a2).
Example(The alternating version)Original
Evaluate Σ2=n=1∑∞n2+4(−1)n.
The set-up is identical, with κalt(z)=πcscπz in place of κ. Using csc(iu)=1/sin(iu)=1/(isinhu)=−icschu,
Check.csch2π=1/sinh2π=0.0037348985, so Σ2=0.0029333824−0.125=−0.1220666176. Direct numerical summation gives −0.1220666176. The sign is right: the first term −1/5 dominates. Note how the two answers differ — coth for the plain sum, csch for the alternating one — which is the whole effect of swapping the kernel.
Example(A pole at an integer: ∑n−2=π2/6)Original
Take f(z)=1/z2, whose only pole, a double one, sits at the integer z=0. Then zf(z)=1/z→0, so the contour estimate of Step 3 applies, and the corrected form of the statement says that all residues of κ(z)/z2 sum to zero.
Residues at n=0. These are f(n)=1/n2, as before.
Residue at 0. The Laurent series of the cotangent about the origin is obtained from cosπz=1−21π2z2+O(z4) and sinπz=πz(1−61π2z2+O(z4)):
Check.π2/6=1.6449340668, the known value of ζ(2). The same computation with f(z)=1/z4 needs the next coefficient of the cotangent expansion, −π4z3/45, and yields ζ(4)=π4/90; carrying the expansion further gives every ζ(2m), each a rational multiple of π2m. Replacing κ by κalt in the same computation gives ∑n≥1(−1)n−1n−2=π2/12, since πcscπz=1/z+π2z/6+O(z3) has residue +π2/6 against 1/z2.
Application(Matsubara sums, lattice energies and Casimir forces)Original
Matsubara sums. In finite-temperature quantum field theory and in many-body perturbation theory, every diagram carries a sum over discrete imaginary frequencies ωn=2πn/β (bosonic) or (2n+1)π/β (fermionic). These are converted into contour integrals by exactly this device, with coth(βz/2) and tanh(βz/2) — the kernels whose poles sit at the Matsubara frequencies with residue 2/β — playing the role of πcotπz. The Bose and Fermi distribution functions that appear in the final answer are nothing but these kernels, and the frequency sum becomes a few residues at the poles of the propagators.
Lattice sums and Madelung constants. The electrostatic energy of a one-dimensional ionic chain is ∑n=0(−1)n/∣n∣-type, and its screened or regularised versions are sums of the form treated here. The alternating kernel is the one that encodes the alternation of charge, and the closed forms in csch reproduce the exponential screening.
Casimir energies. Mode sums ∑nωn over the standing waves in a cavity are regularised and evaluated by writing them as contour integrals with a cot-type kernel; the finite part, after the divergent pieces are identified as residues at the origin, is the Casimir energy. The appearance of ζ(2m) and hence of powers of π in such results traces back to the cotangent expansion used in the third example.
Closed forms for sums that resist elementary methods.∑n≥1(n2+a2)−1=(π/2a)cothπa−1/2a2, derived above, is the workhorse formula behind the summation of Lorentzian lineshapes over an equally spaced comb — a frequency comb, a set of cavity modes, or the images in a periodic Green's function — and it converts a slowly converging sum into a function that can be differentiated and expanded.
The integral of f′/f around a contour counts the zeros minus the poles of f inside it, each with multiplicity; if one function dominates another on the contour, the two have the same number of zeros inside, which locates the roots of polynomials without solving them.
What it means. Zeros and poles are invisible to a contour integral of f itself, but very visible to an integral of f′/f: that function has a simple pole at every zero and every pole of f, with residue the multiplicity, counted positively for zeros and negatively for poles. So one integral counts them. Rouche's theorem turns the count into a practical tool: it says the count is stable under any perturbation smaller in modulus than the function itself on the contour, so a complicated function can be counted by replacing it with the single term that dominates.
A zero of f of order h is counted h times, and a pole of order h is counted h times, in the lists (aj) of zeros and (bk) of poles below. Write
Nf(γ)=j∑n(γ,aj),Pf(γ)=k∑n(γ,bk),
the winding numbers (The winding number) weighting each point. When γ is a positively oriented simple closed contour, every n(γ,⋅) is 1 inside and 0 outside, so Nf and Pf are simply the number of zeros and of poles enclosed, with multiplicity.
Let f be meromorphic on Ω with zeros aj and poles bk, and let γ be a cycle in Ω, homologous to zero in Ω, avoiding all the aj and bk. Then the sums below are finite and
2πi1∮γf(z)f′(z)dz=Nf(γ)−Pf(γ).
Classical form. If γ is a positively oriented simple closed contour with γ and its interior in Ω, and f has neither zero nor pole on γ, then
2πi1∮γf(z)f′(z)dz=N−P,
where N and P count the zeros and poles inside, with multiplicity.
Why "argument". The left-hand side equals n(Γ,0), the winding number about the origin of the image cycle Γ=f∘γ. Equivalently, writing Δγ[argf] for the total change in argf(z) as z traverses γ once,
N−P=2π1Δγ[argf].
Weighted form. If g is analytic on Ω, then under the same hypotheses
2πi1∮γg(z)f(z)f′(z)dz=j∑n(γ,aj)g(aj)−k∑n(γ,bk)g(bk).Theorem(Rouche's theorem)AdaptedAdapted from Ahlfors (1979), Ch. 4, §5.2, Corollary, p. 153
Take a cycle γ in Ω which is homologous to zero there and whose index n(γ,z) is either 0 or 1 at every z off γ. Let f,g be analytic on Ω and suppose that at each point z of γ
∣f(z)−g(z)∣<∣f(z)∣.
Write Z(h) for the number of zeros of h that γ encloses, each repeated according to its order. The conclusion is that Z(f)=Z(g).
Arfken's form. Let F,G be analytic on and inside a simple closed contour C with ∣G∣<∣F∣ at every point of C. Then Z(F)=Z(F+G). This is the same assertion: put f=F and g=F+G, so that f−g=−G.
Idea of the proofOriginal
Take the logarithmic derivative. If f(z)=(z−a)hφ(z) with φ analytic and non-zero near a, then
ff′=z−ah+φφ′,
and the second term is analytic near a. So f′/f has a simple pole at a with residue h — and the same computation with −h in place of h shows the residue at a pole of order h is −h. The residue theorem then delivers the count.
Rouche's theorem is the argument principle applied to the ratio. The hypothesis says ∣g/f−1∣<1 on γ, so the image of γ under F=g/f lies in the disc of centre 1 and radius 1, which misses the origin. A cycle confined to such a disc has winding number 0 about the origin, so F has as many zeros as poles inside — and its zeros are those of g, its poles those of f.
Because φ is analytic and non-zero near z0, the last term is analytic there. Hence f′/f has a simple pole at z0 with
z=z0Resff′=μ,
that is, +h at a zero of order h and −h at a pole of order h. At any other point of Ω, f is analytic and non-zero, so f′/f is analytic.
Step 2: apply the residue theorem. The zeros and poles of f have no accumulation point in Ω: poles are isolated by the definition of meromorphic, and zeros are isolated unless f≡0, which is excluded since f has no zero on γ. So f′/f satisfies the hypotheses of the residue theorem on Ω with singular set the zeros and poles of f, and γ is a cycle in the complement of that set, homologous to zero in Ω. Therefore (The residue theorem)
the zeros and poles being listed with multiplicity so that an order-h point contributes h equal terms rather than a single residue h; the residue theorem also guarantees only finitely many terms are non-zero. The classical form is the case n(γ,⋅)≡1 inside.
Step 3: the argument interpretation. Let Γ=f∘γ, a cycle in C∖{0} since f=0 on γ. Parametrising γ by t and substituting w=f(γ(t)),
By the proof of integrality of the index (The winding number), 2πin(Γ,0) is the total increment of a continuous branch of logw along Γ, whose imaginary part is the total increment of argf along γ; the real part cancels because ∣f∣ returns to its starting value. This gives N−P=Δγ[argf]/2π.
Step 4: the weighted form. With g analytic on Ω, Step 1 gives
g(z)f(z)f′(z)=z−z0μg(z)+g(z)φ(z)φ′(z)
near z0, and the residue of the first term is μg(z0) while the second is analytic. The residue theorem again gives the stated sum.
Step 5: Rouche's theorem. The hypothesis ∣f−g∣<∣f∣ on γ forces ∣f∣>0 there, hence also ∣g∣≥∣f∣−∣f−g∣>0: neither function vanishes on γ. Dividing the hypothesis by ∣f(z)∣,
f(z)g(z)−1<1on γ.
Set F=g/f, meromorphic on Ω with zeros the zeros of g and poles the zeros of f (a common zero of f and g cancels, affecting both counts equally and so not the conclusion). The image cycle F∘γ lies in the open disc ∣w−1∣<1. That disc omits 0 and its complement is connected and unbounded, so 0 lies in the unbounded component of the complement of the image, whence n(F∘γ,0)=0 (The winding number). By Steps 2 and 3 applied to F,
0=n(F∘γ,0)=NF(γ)−PF(γ)=Ng(γ)−Nf(γ).
Since n(γ,⋅)∈{0,1}, Ng(γ) and Nf(γ) are the numbers of zeros of g and of f enclosed by γ, and they are equal. □
Remark(Where the hypotheses are used)Original
No zeros or poles on the contour. Both f′/f and argf are undefined there, and the count genuinely jumps as a zero crosses γ. This is the only hypothesis that fails in practice, and it fails exactly when a root sits on the test circle.
Which function dominates, and why the inequality is strict. The bound is ∣f−g∣<∣f∣, with the dominant function on the right; swapping the roles is a different hypothesis. On ∣z∣=1 with f(z)=z and g(z)=1 we have ∣f−g∣=∣z−1∣, which reaches 2>1 at z=−1: the hypothesis fails, and so does the conclusion, f having one zero inside and g none. Equality cannot be permitted even at a single point, because ∣g/f−1∣=1 allows g(z)=0 there, and a zero sitting on γ leaves "the number of zeros enclosed" undefined. Concretely, f(z)=1 and g(z)=1−z satisfy ∣f−g∣=∣z∣=1=∣f∣ at every point of ∣z∣=1, and the zero of g lies exactly on the contour.
Multiplicity is unavoidable. The integral cannot distinguish a double zero from two simple ones. Rouche's theorem therefore tells you how many roots are in a region, never where they are; locating them needs either a smaller region or a different method.
The {0,1} condition on the index in Rouche. Without it the conclusion is about weighted counts Nf(γ), not about numbers of zeros, and "the same number of zeros enclosed" would be the wrong phrase.
Meromorphic, not merely analytic. The argument principle is stated for meromorphic f precisely so that it can count poles too. For analytic f, P=0 and the statement reduces to Ahlfors's earlier formula (31) on p. 130 counting zeros alone.
Example(Trapping the roots of a quintic)Original
Let p(z)=z5+3z+1. We locate its five roots without solving.
How many in ∣z∣<1? On the circle ∣z∣=1 take f(z)=3z as the dominant part and g(z)=p(z)=3z+(z5+1). Then
∣f(z)−g(z)∣=∣z5+1∣≤∣z∣5+1=2<3=∣3z∣=∣f(z)∣.
The inequality is strict at every point of the circle, so Rouche applies: p has as many zeros in ∣z∣<1 as 3z, namely one.
How many in ∣z∣<2? Now the leading term dominates. On ∣z∣=2 take f(z)=z5:
∣p(z)−z5∣=∣3z+1∣≤3⋅2+1=7<32=∣z∣5.
So p has five zeros in ∣z∣<2: all of them, consistent with the degree.
Conclusion. One root lies in ∣z∣<1 and the remaining four in the annulus 1<∣z∣<2. Since p has real coefficients, the non-real roots come in conjugate pairs; the single small root must therefore be real, and indeed p(0)=1>0 while p(−1)=−3<0, so it lies in (−1,0) by the intermediate value theorem.
Check. The actual roots are −0.3319890, 1.0050669±0.9372592i and −0.8390724±0.9438516i, with moduli 0.3320, 1.3743 (twice) and 1.2629 (twice) — exactly one inside the unit circle, four in the annulus. Evaluating the argument-principle integral numerically confirms the counts: 2πi1∮∣z∣=1p′/pdz=1.000000 and 2πi1∮∣z∣=2p′/pdz=5.000000.
Example(Counting a pole against a zero)Original
Let
f(z)=(z−i)3z2(z−3),
meromorphic on C, and let γ be the circle ∣z∣=2, positively oriented.
Inventory. Inside γ: a zero of order 2 at z=0 and a pole of order 3 at z=i (since ∣i∣=1<2). Outside: the simple zero at z=3. Nothing lies on γ, since ∣0∣,∣i∣<2<∣3∣. So N=2, P=3 and the argument principle predicts
2πi1∮∣z∣=2f(z)f′(z)dz=2−3=−1.
Direct confirmation by the logarithmic derivative. Because f is a product of powers,
ff′=z2+z−31−z−i3,
obtained by differentiating logf=2logz+log(z−3)−3log(z−i) term by term (valid locally; only the derivative is needed). Integrating over ∣z∣=2 and using 2πi1∮dz/(z−a)=n(γ,a):
2⋅1+1⋅0−3⋅1=−1,
since n(γ,0)=n(γ,i)=1 and n(γ,3)=0. A numerical integration returns −1.0000000.
Reading it as a phase. Equivalently, as z runs once counterclockwise around ∣z∣=2, the point f(z) winds once clockwise about the origin. The excess of pole order over zero order shows up as a reversal of the direction of winding — which is exactly what a Nyquist plot reports.
Application(Stability, root finding and mode counting)Original
Nyquist stability. A linear feedback system with open-loop transfer function G(s) is closed-loop stable when 1+G(s) has no zeros in the right half-plane. Taking γ to be the boundary of that half-plane (the imaginary axis closed by a large semicircle), the argument principle gives N−P=n(Γ,−1), where Γ is the Nyquist plot of G and P is the known number of right-half-plane poles of G. The engineer counts encirclements of −1 and reads off N; stability is N=0. No root of the characteristic polynomial is ever computed.
Locating roots in a region. Rouche's theorem is the standard way to trap polynomial roots in annuli, as in the example above and in the treatment of z4+z+1 by Riley, Hobson and Bence (pp. 880-882), who combine a quadrant-by-quadrant argument count with two applications of Rouche to show that each quadrant contains one root and all four lie between ∣z∣=2/3 and ∣z∣=3/2. Numerical root finders use the same integral as a reliable way to decide how many roots a candidate region contains before subdividing.
Counting modes and bound states. For a dispersion relation D(ω)=0 or a Jost function f(k), the number of bound states or unstable modes in a region of the complex plane is the number of zeros there, and the contour integral of D′/D counts them robustly even when the zeros are nearly degenerate. Levinson's theorem, relating the total scattering phase shift to the number of bound states, is an argument principle in this form: a change in total argument counts zeros.
Perturbation stability. Rouche's theorem says a zero count cannot change under a perturbation that is small compared with the function on the boundary. This is the precise sense in which "a simple root moves continuously and does not disappear" when a model is perturbed — and the sense in which a root can escape only by reaching the boundary where the dominance hypothesis fails.
The index of a closed curve about a point is the integral of dz/(z−a) divided by 2πi; it is an integer, constant on each complementary region and zero far away, unchanged by continuous deformation, and it is the weight with which each point enters the general form of Cauchy's theorem.
What it means. "The contour encircles the pole twice" is an intuitive statement that has to be made into a number before it can appear in a theorem. The number is the winding number, and it is nothing but the integral ∮dz/(z−a) in disguise. Its three properties do all the work: it is always an integer, it only changes when the point crosses the curve, and it is blind to continuous deformations of the curve that avoid the point. Those properties are what turn Cauchy's theorem from a statement about discs into a statement about arbitrary regions and arbitrary closed curves.
Definition(Chains and cycles)AdaptedAdapted from Ahlfors (1979), Ch. 4, §4.1, pp. 137-138
A chain in an open set Ω is a formal integer combination γ=∑jmjγj of piecewise differentiable arcs γj in Ω, with the integral defined additively:
∮γf(z)dz=j∑mj∫γjf(z)dz.
Two chains are regarded as the same when they give the same integral for every f; permuting, subdividing, fusing and reparametrising arcs, and cancelling an arc against its reverse, all leave a chain unchanged. A chain is a cycle if it can be written as a sum of closed curves, which happens exactly when in any representation the initial and terminal points of the constituent arcs match up in pairs.
Definition(Index of a point with respect to a cycle)AdaptedAdapted from Ahlfors (1979), Ch. 4, §2.1, p. 115
Take a point a and a cycle γ that avoids it. Then the number
n(γ,a)=2πi1∮γz−adz
is called the index of a relative to γ, or its winding number. Two immediate consequences of the additivity of contour integrals: running the cycle backwards flips the sign, n(−γ,a)=−n(γ,a), and the index adds over cycles, n(γ1+γ2,a)=n(γ1,a)+n(γ2,a).
Definition(Basic properties of the index)AdaptedAdapted from Ahlfors (1979), Ch. 4, §2.1, Lemma 1 and properties (i)-(ii), pp. 115-116
Let γ be a cycle and a∈/γ.
Integrality.n(γ,a)∈Z.
Local constancy. As a function of a, the index is constant on each connected component of the complement C∖γ.
Vanishing at infinity. The complement has exactly one unbounded component, and n(γ,a)=0 there; in particular n(γ,a)=0 whenever γ lies inside a circle and a lies outside it.
Multiplicativity. If γ is a closed curve t↦γ(t) which is a pointwise product γ(t)=u(t)v(t) of two closed curves avoiding 0, then n(γ,0)=n(u,0)+n(v,0).
Definition(Homotopy invariance)AdaptedAdapted from Ahlfors (1979), Ch. 8, §1.5, Definition 2, p. 292; Ahlfors (1979), Ch. 4, §2.1, property (ii), p. 116
Let a∈C and let γ0,γ1:[0,1]→C∖{a} be closed, piecewise differentiable curves that are homotopic in C∖{a} through closed curves: that is, there is a continuous H:[0,1]×[0,1]→C∖{a} with H(t,0)=γ0(t), H(t,1)=γ1(t) and H(0,u)=H(1,u) for all u (Connectedness and simple connectivity). Then
n(γ0,a)=n(γ1,a).
In particular a closed curve that can be contracted to a point within C∖{a} has index 0 about a.
Definition(Simple connectivity and homology)AdaptedAdapted from Ahlfors (1979), Ch. 4, §4.2, Definition 1 and Thm 14, pp. 139-140; Ahlfors (1979), Ch. 4, §4.3, Definition 2, p. 141
Call a region Ωsimply connected when the set (C∪{∞})∖Ω — the complement taken in the extended plane, so that the point at infinity is included — is a connected set. Ahlfors's Thm 14 shows this is the same as requiring n(γ,a)=0 for every cycle γ in Ω and every a∈/Ω.
A cycle γ in an open set Ω is homologous to zero in Ω, written γ∼0(modΩ), if
n(γ,a)=0for every a∈C∖Ω.
Write γ1∼γ2 for γ1−γ2∼0. Homologies may be added and subtracted, and γ∼0(modΩ) implies γ∼0(modΩ′) for every larger open set Ω′⊇Ω.
A warning on terminology. The definition adopted here is not the standard one, as Ahlfors himself points out (p. 139). Most texts define simple connectivity by contractibility of closed curves. For plane regions the two conditions agree; the version used here is restricted to two real dimensions, but it reaches the integration theorems with less work, which is why Ahlfors prefers it and why we follow him.
Definition(The general (homology) form of Cauchy's theorem)AdaptedAdapted from Ahlfors (1979), Ch. 4, §4.4, Thm 15 and Corollary 1, p. 141; Ahlfors (1979), Ch. 4, §5.1, p. 149
Let f be analytic on an open set Ω.
Cauchy's theorem. If γ is a cycle in Ω with γ∼0(modΩ), then
∮γf(z)dz=0.
Corollary. If Ω is simply connected, this holds for every cycle γ in Ω; equivalently, every analytic function on a simply connected region has a single-valued analytic antiderivative there.
Cauchy's integral formula. If γ∼0(modΩ) and a∈Ω is not on γ, then
n(γ,a)f(a)=2πi1∮γz−af(z)dz.
Idea of the proofOriginal
Integrality is the only point needing a real argument, and the argument is a trick: instead of trying to define arg(z−a) continuously along the curve, integrate first and exponentiate afterwards. Let h(t) be the partial integral up to time t. Differentiating e−h(t)(γ(t)−a) gives zero, so this quantity is constant along the curve; since the curve closes, eh(1)=1, which says exactly that h(1) is an integer multiple of 2πi.
Local constancy is a one-line application of Cauchy's theorem to the function (z−a)−1−(z−b)−1, which has the single-valued antiderivative logz−bz−a off the segment joining a to b. Vanishing at infinity then follows by pushing a out beyond a circle containing γ and using the ML estimate.
Homotopy invariance combines the two: a continuous deformation moves the index continuously, and an integer-valued continuous function of a real parameter is constant.
ProofAdaptedAdapted from Ahlfors (1979), Ch. 4, §2.1, pp. 115-116
It is enough to prove parts 1-3 for a single closed curve, since both sides of each assertion are additive in the cycle. Let γ:[0,1]→C be closed and piecewise differentiable with γ(t)=a for all t.
Part 1: integrality. Define
h(t)=∫0tγ(s)−aγ′(s)ds,t∈[0,1],
which is continuous on [0,1] and satisfies h′(t)=γ′(t)/(γ(t)−a) wherever γ′ is continuous, i.e. off a finite set. Consider
Φ(t)=e−h(t)(γ(t)−a).
Away from that finite set,
Φ′(t)=e−h(t)(γ′(t)−h′(t)(γ(t)−a))=0,
by the formula for h′. Since Φ is continuous on [0,1] and has vanishing derivative off finitely many points, it is constant. Hence
eh(t)=γ(0)−aγ(t)−a,
using h(0)=0. The curve is closed, γ(1)=γ(0), so eh(1)=1 and therefore h(1)∈2πiZ. As n(γ,a)=h(1)/2πi, the index is an integer. (The naive argument — that log∣z−a∣ returns to its value and arg(z−a) changes by a multiple of 2π — is not a proof until arg(z−a) has been given a single-valued meaning along γ, which is the difficulty this computation avoids.)
Part 2: local constancy. Any two points of the same component of C∖γ can be joined by a polygonal path missing γ, so it suffices to show n(γ,a)=n(γ,b) whenever the straight segment [a,b] does not meet γ. Off that segment the quotient (z−a)/(z−b) is never zero and never a negative real number: if it were ≤0 then z would lie on the segment between a and b. Hence the principal branch of logz−bz−a is single-valued and analytic on C∖[a,b], with derivative
dzdlogz−bz−a=z−a1−z−b1.
Since γ is a closed curve in C∖[a,b] and the integrand is an exact differential there, the fundamental theorem for contour integrals (Contour integrals) gives
∮γ(z−a1−z−b1)dz=0,
that is n(γ,a)=n(γ,b).
Part 3: vanishing at infinity. As a point set γ is compact, so it lies in some disc ∣z∣<ρ; the set ∣z∣>ρ is connected and lies in the complement, so the complement has exactly one unbounded component. For ∣a∣>ρ the ML estimate (Contour integrals) gives
∣n(γ,a)∣≤2π1⋅∣a∣−ρL(γ)∣a∣→∞0,
where L(γ) is the length of γ. An integer of modulus less than 1 is 0, so n(γ,a)=0 for all sufficiently large ∣a∣; by part 2 it is then 0 throughout the unbounded component.
Part 4: multiplicativity. With γ=uv pointwise, γ′=u′v+uv′ off finitely many points, so
γγ′=uvu′v+uv′=uu′+vv′,
and integrating over [0,1] and dividing by 2πi gives n(γ,0)=n(u,0)+n(v,0). □
Proof(Homotopy invariance)Original
Let H be the homotopy of the statement. We first reduce to a uniform estimate. The image K=H([0,1]2) is compact and avoids a, so
ϵ=(t,u)min∣H(t,u)−a∣>0,
and H is uniformly continuous, so there is η>0 with ∣H(t,u)−H(t,u′)∣<ϵ/2 whenever ∣u−u′∣<η.
Step 1: a comparison lemma.If σ and τ are closed piecewise differentiable curves with ∣σ(t)−a∣≥δ and ∣σ(t)−τ(t)∣<δ for all t and some δ>0, then n(σ,a)=n(τ,a).
Both curves avoid a, since ∣τ−a∣≥∣σ−a∣−∣σ−τ∣>0. Put w(t)=(τ(t)−a)/(σ(t)−a), a closed piecewise differentiable curve. Then
∣w(t)−1∣=∣σ(t)−a∣∣τ(t)−σ(t)∣<δδ=1,
so w lies in the open disc ∣w−1∣<1, which does not contain 0. Since the complement of that disc is connected and unbounded, 0 lies in the unbounded component of C∖w, so n(w,0)=0 by part 3 of the basic properties. Writing τ(t)−a=(σ(t)−a)w(t) and applying multiplicativity (part 4) to the translated curves,
n(τ,a)=n(σ,a)+n(w,0)=n(σ,a).
Step 2: polygonal approximants. The intermediate curves H(⋅,u) need not be piecewise differentiable, so replace each by a polygon. By uniform continuity of H on the compact square, choose N∈N with
∣H(t,u)−H(t′,u)∣<ϵ/16whenever ∣t−t′∣≤1/N,
and let Pu be the closed polygon that passes through H(k/N,u) for k=0,…,N, parametrised affinely on each [k/N,(k+1)/N]. It is closed because H(0,u)=H(1,u). For t in that subinterval, Pu(t) lies on the segment joining two points at distance less than ϵ/16, hence within ϵ/16 of H(k/N,u), so
Step 3: step through the deformation. By uniform continuity again, choose m with ∣H(t,u)−H(t,u′)∣<ϵ/4 whenever ∣u−u′∣≤1/m, and put ui=i/m. For consecutive i,
Step 1 with δ=ϵ/2 therefore gives n(Pui,a)=n(Pui+1,a) for every i, so all the Pui have the same index.
Step 4: the two ends. Since ∣γ0(t)−Pu0(t)∣=∣H(t,0)−P0(t)∣<ϵ/8<ϵ/2 and ∣γ0(t)−a∣≥ϵ>ϵ/2, Step 1 gives n(γ0,a)=n(Pu0,a); likewise n(γ1,a)=n(Pum,a). Combining with Step 3, n(γ0,a)=n(γ1,a).
For the last sentence of the statement: a curve contractible in C∖{a} is homotopic there to a constant curve, whose index is 0 because the integrand integrates over a degenerate path. □
Remark(Where the hypotheses are used)Original
a∈/γ is indispensable. If a lies on the curve the integral diverges and no index is defined. This is why the residue theorem requires the contour to miss the singularities, and why a pole on the contour must be handled separately (Principal values and indented contours).
"Through closed curves" in the homotopy. The deformation must keep the curve closed at every stage; otherwise the index can change. A homotopy of arcs with fixed endpoints is a different notion and does not apply to a cycle.
The homotopy must avoid a. Deforming the unit circle to a point through the origin changes its index about 0 from 1 to 0. Step 1 fails precisely because ϵ=0.
Homology is weaker than homotopy. A cycle homologous to zero need not be null-homotopic. In the twice-punctured plane Ω=C∖{0,1} the commutator cycle built from a loop around 0 and a loop around 1 has index 0 about both punctures, so it is homologous to zero and Cauchy's theorem applies to it; it is not contractible in Ω. This is why the integration theorems are stated with homology, which is exactly what they need: Ahlfors credits the observation to Emil Artin.
Simple connectivity is about the complement in the extended plane. Ahlfors stresses this (p. 139): a parallel strip is simply connected, although its complement in the finite plane is disconnected, while the exterior of a circle is not, because its complement consists of a closed disc and the separate point ∞.
Example(Computing an index three ways)Original
(a) A clockwise ellipse. Let γ(t)=eit+2e−it for t∈[0,2π]. Writing it in real terms, γ(t)=3cost−isint, so the image is the ellipse (x/3)2+y2=1 traversed clockwise (at t=0 it is at 3, and a moment later the imaginary part is negative). Hence n(γ,a)=−1 for every a inside the ellipse and 0 outside. Direct numerical integration confirms this: n(γ,0)=−1.000000, n(γ,2)=−1.000000 (the point 2 is inside, since (2/3)2<1), and n(γ,4)=0 to twelve digits. The lesson is that the sign of the index is not decided by the sign in the exponent e+it: the dominant term is 2e−it.
(b) A doubly wound curve, by factorisation. Let σ(t)=e2it+21e−it. Factor
Both factors are closed curves avoiding 0: ∣v(t)−1∣=21, so v lies on a circle of radius 21 about 1 and 0 is in the unbounded component of its complement, giving n(v,0)=0; and n(u,0)=2 because ∮u′/udt=∫02π2idt=4πi. By multiplicativity, n(σ,0)=2+0=2. Numerically, 2.0000000. For the point a=2: ∣σ(t)∣≤1+21=23<2, so 2 lies in the unbounded component of the complement and n(σ,2)=0, again confirmed numerically.
(c) A contour integral read backwards. Take for γ the positively oriented perimeter of the square whose four corners are ±2±2i, and put a=1+i. Rather than integrate, deform: γ is homotopic in C∖{a} to the circle ∣z−a∣=1/2 traversed once counterclockwise, by the straight-line homotopy from the square to that circle, which stays in the region between them and so avoids a. Hence n(γ,a)=1 by homotopy invariance, and
∮γz−(1+i)ezdz=2πie1+i
by the integral formula in part 3 above. Numerically e1+i=1.4686939399+2.2873552872i, so the right-hand side is −14.3718771327+9.2280761840i; integrating along the four sides of the square numerically returns the same ten digits.
Application(Topological charge and the Nyquist criterion)Original
Vortices and topological charge. In a superfluid, a superconductor or a liquid crystal, the order parameter is a complex field ψ. Around a closed loop on which ψ=0, the phase must return to itself up to a multiple of 2π, and the integer
2πi1∮Cψ∇ψ⋅dℓ
is the winding number of the image curve about the origin: the topological charge of the vortices enclosed. Integrality is why circulation in a superfluid is quantised, and homotopy invariance is why the charge cannot change under any continuous evolution of the field that keeps ψ=0 on the loop. A vortex can only be destroyed by letting ψ pass through zero — exactly the forbidden deformation of the remark above.
The Nyquist stability criterion. For a feedback loop with open-loop transfer function G(s), closed-loop stability is governed by the zeros of 1+G(s) in the right half-plane. Taking γ to be a large contour enclosing that half-plane, the number of encirclements of the point −1 by the image curve G(γ) — which is a winding number — together with the known number of right-half-plane poles of G, decides stability (Argument principle and Rouche's theorem). The engineering practice of "counting how many times the Nyquist plot encircles −1" is a computation of n(G∘γ,−1), and its robustness under small modelling errors is homotopy invariance.
Degree and root counting. The winding number of p(γ) about 0, for γ a large circle and p a polynomial, equals degp; this is one route to the fundamental theorem of algebra and the basis of numerical algorithms that count roots in a prescribed region by integrating p′/p along its boundary.