Technopedia

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Complex integration

Complex integration integrates holomorphic functions along curves (contours) in the complex plane. Because holomorphic functions are rigid, contour integrals obey powerful laws: they are independent of the path, they recover the function’s values and derivatives from boundary data (Cauchy’s formula), and they reduce to a finite sum of local data at isolated singularities (the residue theorem). These tools evaluate otherwise intractable real integrals and underpin Laplace-transform inversion in Differential equations and spectral methods in Functional analysis.

Authoritative sources

Contour integrals and Cauchy’s integral theorem

Definition. If ff is continuous on a curve γ:[a,b]→C\gamma: [a,b] \to \mathbb{C} (a piecewise C1C^1 path), its contour integral is

∫γf(z) dz=∫abf(γ(t)) γ′(t) dt.\int_\gamma f(z)\, dz = \int_a^b f(\gamma(t))\, \gamma'(t)\, dt.

Cauchy’s integral theorem. If ff is holomorphic on a simply connected open set UU and γ\gamma is a closed contour in UU, then

∫γf(z) dz=0.\int_\gamma f(z)\, dz = 0.

Details

What a contour integral means

A contour integral is the real line integral of a complex-valued function. Writing f=u+ivf = u + iv and z=x+iyz = x + iy, we have dz=dx+i dydz = dx + i\, dy, so

∫γf dz=∫γ(u dx−v dy)+i∫γ(v dx+u dy).\int_\gamma f\, dz = \int_\gamma (u\, dx - v\, dy) + i \int_\gamma (v\, dx + u\, dy).

It is a genuine complex number, not a magnitude.

Why the integral vanishes on closed loops

The key is the Cauchy–Riemann equations. For a small rectangle, the integral of f dzf\, dz around its boundary can be written in terms of ux,uy,vx,vyu_x, u_y, v_x, v_y. The Cauchy–Riemann equations make the contributions from opposite edges cancel. Tiling a simply connected region into small rectangles, all interior edges cancel pairwise, leaving only the outer boundary — and the sum tends to zero as the mesh refines. Hence the integral over any closed contour is zero.

The simply connected hypothesis is essential: it rules out “holes.” Around a hole the integral need not vanish (see the residue theorem below).

Path independence

A direct consequence: in a simply connected domain, the integral of a holomorphic function depends only on the endpoints, not the path. This lets us define an antiderivative F(z)=∫z0zf(ζ) dζF(z) = \int_{z_0}^{z} f(\zeta)\, d\zeta, which is holomorphic with F′=fF' = f.

z0 z1 In a simply connected domain, the integral of a holomorphic f is the same along any two paths.
Example problem

Problem. Evaluate ∮∣z∣=11z2 dz\oint_{|z|=1} \frac{1}{z^2}\, dz.

Solution

The function 1/z21/z^2 is holomorphic on C∖{0}\mathbb{C} \setminus \{0\} and has an antiderivative −1/z-1/z on the punctured plane. Since the contour ∣z∣=1|z| = 1 is closed and −1/z-1/z is single-valued, the integral is 00. (Equivalently, the residue of 1/z21/z^2 at 00 is 00, so the residue theorem gives 00.) ∎

Cauchy’s integral formula

Theorem. If ff is holomorphic on a simply connected open set containing a closed disk D(z0,r)‾\overline{D(z_0, r)} and γ\gamma is the positively oriented boundary of that disk, then for ∣z−z0∣<r|z - z_0| < r,

f(z)=12πi∫γf(ζ)ζ−z dζ.f(z) = \frac{1}{2\pi i} \int_\gamma \frac{f(\zeta)}{\zeta - z}\, d\zeta.

Differentiating under the integral gives the derivatives:

f(n)(z)=n!2πi∫γf(ζ)(ζ−z)n+1 dζ.f^{(n)}(z) = \frac{n!}{2\pi i} \int_\gamma \frac{f(\zeta)}{(\zeta - z)^{n+1}}\, d\zeta.

Details

The idea

Cauchy’s formula expresses the value of ff at an interior point purely in terms of its values on the surrounding contour. The kernel 1/(ζ−z)1/(\zeta - z) is holomorphic in ζ\zeta except at ζ=z\zeta = z, so the integral is a “local probe” of ff at zz. This is the computational engine of the subject: it turns point values into contour integrals and vice versa.

Worked example

Take f(ζ)=ζ2f(\zeta) = \zeta^2 and γ\gamma the unit circle. Then

f(0)=12πi∮∣z∣=1z2z−0 dz=12πi∮∣z∣=1z dz=0,f(0) = \frac{1}{2\pi i} \oint_{|z|=1} \frac{z^2}{z - 0}\, dz = \frac{1}{2\pi i} \oint_{|z|=1} z\, dz = 0,

consistent with f(0)=0f(0) = 0. For a nonzero point, say f(1/2)f(1/2):

f(1/2)=12πi∮∣z∣=1z2z−1/2 dz.f(1/2) = \frac{1}{2\pi i} \oint_{|z|=1} \frac{z^2}{z - 1/2}\, dz.

Since z2/(z−1/2)=z+1/2+1/4z−1/2z^2/(z - 1/2) = z + 1/2 + \frac{1/4}{z - 1/2}, only the last term contributes (the others have antiderivatives), giving f(1/2)=1/4=(1/2)2f(1/2) = 1/4 = (1/2)^2.

Why it matters

Cauchy’s formula immediately implies that holomorphic functions are C∞C^\infty and real-analytic (differentiate as many times as you like), and it yields the Cauchy estimates ∣f(n)(z0)∣≤n!M/rn|f^{(n)}(z_0)| \leq n! M / r^n (where M=max⁡γ∣f∣M = \max_\gamma |f|), which control the growth of the Taylor coefficients.

Example problem

Problem. Let ff be holomorphic on ∣z∣≤2|z| \leq 2 with ∣f(z)∣≤10|f(z)| \leq 10 there. Bound ∣f′′(0)∣|f''(0)|.

Solution

By the Cauchy estimate with n=2n = 2, r=2r = 2, M=10M = 10: ∣f′′(0)∣≤2!⋅10/22=20/4=5|f''(0)| \leq 2! \cdot 10 / 2^2 = 20/4 = 5. ∎

Liouville’s theorem and the fundamental theorem of algebra

Liouville’s theorem. Every bounded entire function (holomorphic on all of C\mathbb{C}) is constant.

Fundamental theorem of algebra (consequence). Every non-constant polynomial p(z)p(z) of degree n≥1n \geq 1 has at least one complex root.

Details

Proof of Liouville’s theorem

Let ff be entire and bounded, ∣f(z)∣≤M|f(z)| \leq M. Fix z0z_0 and apply Cauchy’s formula for the first derivative on a disk of radius rr:

∣f′(z0)∣≤Mr.|f'(z_0)| \leq \frac{M}{r}.

Letting r→∞r \to \infty (possible because ff is entire) gives f′(z0)=0f'(z_0) = 0. Since z0z_0 was arbitrary, f′≡0f' \equiv 0, so ff is constant.

From Liouville to the fundamental theorem of algebra

Let pp be a non-constant polynomial with no root. Then g(z)=1/p(z)g(z) = 1/p(z) is entire. As ∣z∣→∞|z| \to \infty, ∣p(z)∣→∞|p(z)| \to \infty, so ∣g(z)∣→0|g(z)| \to 0; hence gg is bounded. By Liouville, gg is constant, so pp is constant — a contradiction. Thus pp has a root.

Example problem

Problem. Show that f(z)=ezf(z) = e^z is not constant but is not bounded, and explain why this does not contradict Liouville’s theorem.

Solution

ez=ex+iy=exeiye^z = e^{x+iy} = e^x e^{iy} has modulus exe^x, which is unbounded as x→+∞x \to +\infty. Liouville’s theorem requires boundedness, which eze^z lacks, so there is no contradiction. (Indeed eze^z is a non-constant entire function, exactly the kind Liouville’s hypothesis excludes.) ∎

The residue theorem

Definition. If ff is holomorphic in a punctured neighborhood of an isolated singularity aa, the residue Res⁡(f,a)\operatorname{Res}(f, a) is the coefficient a−1a_{-1} of (z−a)−1(z-a)^{-1} in the Laurent expansion of ff about aa. For a pole of order mm,

Res⁡(f,a)=1(m−1)!lim⁡z→adm−1dzm−1[(z−a)mf(z)].\operatorname{Res}(f, a) = \frac{1}{(m-1)!} \lim_{z \to a} \frac{d^{m-1}}{dz^{m-1}} \big[(z-a)^m f(z)\big].

For a simple pole (m=1m = 1), Res⁡(f,a)=lim⁡z→a(z−a)f(z)\operatorname{Res}(f, a) = \lim_{z \to a} (z-a) f(z).

Residue theorem. If ff is holomorphic inside and on a positively oriented simple closed contour γ\gamma except for isolated singularities a1,…,ana_1, \dots, a_n inside γ\gamma, then

∮γf(z) dz=2πi∑k=1nRes⁡(f,ak).\oint_\gamma f(z)\, dz = 2\pi i \sum_{k=1}^{n} \operatorname{Res}(f, a_k).

Details

Why the residue is the only thing that matters

Near an isolated singularity, ff has a Laurent series ∑n=−∞∞an(z−a)n\sum_{n=-\infty}^{\infty} a_n (z-a)^n. Every term an(z−a)na_n (z-a)^n with n≠−1n \neq -1 has a single-valued antiderivative on the punctured disk, so its integral around a closed loop is zero. Only the a−1/(z−a)a_{-1}/(z-a) term integrates to 2πia−12\pi i a_{-1}. The residue is therefore precisely the “non-cancelling” part of the singularity.

Worked example: a classic real integral

Evaluate ∫−∞∞dx1+x2\int_{-\infty}^{\infty} \frac{dx}{1 + x^2}. Close the contour in the upper half-plane with a large semicircle. The integrand 1/(1+z2)=1/((z−i)(z+i))1/(1+z^2) = 1/((z-i)(z+i)) has a simple pole at z=iz = i (inside) and z=−iz = -i (outside). The residue at ii is

lim⁡z→i(z−i)1(z−i)(z+i)=12i=−i2.\lim_{z \to i} (z-i) \frac{1}{(z-i)(z+i)} = \frac{1}{2i} = -\frac{i}{2}.

The semicircle contribution vanishes as its radius →∞\to \infty (the integrand is O(1/R2)O(1/R^2) on an arc of length πR\pi R). Hence

∫−∞∞dx1+x2=2πi⋅(−i2)=π.\int_{-\infty}^{\infty} \frac{dx}{1+x^2} = 2\pi i \cdot \left(-\frac{i}{2}\right) = \pi.

The general recipe

  1. Identify the singularities of ff and which lie inside the chosen contour.
  2. Compute each residue (simple-pole formula, or the order-mm formula).
  3. Sum them and multiply by 2πi2\pi i.
  4. For real integrals, justify that the added arc contributes zero in the limit.
z = i (inside) z = -i (outside) Closing in the upper half-plane: only the enclosed pole contributes to the integral.
Example problem

Problem. Compute ∮∣z∣=2zz2+1 dz\oint_{|z|=2} \frac{z}{z^2 + 1}\, dz.

Solution

The poles are at z=±iz = \pm i, both inside ∣z∣=2|z| = 2. Both are simple. At z=iz = i: Res⁡=i2i=1/2\operatorname{Res} = \frac{i}{2i} = 1/2. At z=−iz = -i: Res⁡=−i−2i=1/2\operatorname{Res} = \frac{-i}{-2i} = 1/2. Sum =1= 1. Hence the integral is 2πi⋅1=2πi2\pi i \cdot 1 = 2\pi i. ∎

Quiz

Example problem

Problem. Evaluate ∫−∞∞dx(x2+1)2\int_{-\infty}^{\infty} \frac{dx}{(x^2 + 1)^2} using the residue theorem.

Solution

Close in the upper half-plane. The integrand 1/(z2+1)2=1/((z−i)2(z+i)21/(z^2+1)^2 = 1/((z-i)^2 (z+i)^2 has a double pole at z=iz = i (inside) and z=−iz = -i (outside). For the double pole at ii, use the order-m=2m=2 formula with g(z)=(z−i)2f(z)=1/(z+i)2g(z) = (z-i)^2 f(z) = 1/(z+i)^2:

Res⁡(f,i)=ddz1(z+i)2∣z=i=−2(z+i)−3∣z=i=−2(2i)−3=−2/(−8i)=14i=−i4.\operatorname{Res}(f, i) = \left.\frac{d}{dz} \frac{1}{(z+i)^2}\right|_{z=i} = \left. -2 (z+i)^{-3} \right|_{z=i} = -2 (2i)^{-3} = -2 / (-8i) = \frac{1}{4i} = -\frac{i}{4}.

The arc vanishes in the limit (integrand is O(1/R4)O(1/R^4)). Therefore

∫−∞∞dx(x2+1)2=2πi⋅(−i4)=π2.\int_{-\infty}^{\infty} \frac{dx}{(x^2+1)^2} = 2\pi i \cdot \left(-\frac{i}{4}\right) = \frac{\pi}{2}.

∎