Mathematics.

contour integration

Contour Integrals

Complex Analysis30 minDifficulty5 out of 10

You should know: analytic functions

Overview

A contour integral generalizes the ordinary definite integral to integration along a curve C in the complex plane rather than along a real interval. If C is parametrized by z(t) = x(t) + iy(t) for t in [a,b], the contour integral of f along C is defined by substituting the parametrization and integrating with respect to t: ∫_C f(z) dz = ∫_a^b f(z(t)) z'(t) dt. Contour integrals depend in general on the path taken, not just its endpoints, unless f is analytic on a simply connected region containing the path (Cauchy's theorem). Closed contours, denoted with the symbol ∮, are especially important because their values are governed by the singularities enclosed, which is the foundation for the residue theorem.

Intuition

Think of ∫_C f(z) dz as summing up tiny complex products f(z_k)·Δz_k as you walk along the curve C in small steps Δz_k — exactly the Riemann-sum idea from real calculus, except each term is now a complex number, so both direction of travel and the rotating/scaling action of f matter. Unlike real integrals, whose value only depends on the interval's endpoints (via the Fundamental Theorem of Calculus, when an antiderivative exists), a contour integral can depend on the whole shape of the path — walking around a pole via two different routes can give completely different answers, unless the function is analytic everywhere between the two paths.

Formal Definition

Definition

For a smooth curve C: z(t), t ∈ [a,b], and f continuous on C:

∫Cf(z) dz=∫abf(z(t)) z′(t) dt\int_C f(z)\,dz = \int_a^b f(z(t))\,z'(t)\,dt
Definition via parametrization
∮Cf(z) dz\oint_C f(z)\,dz
Notation for integration over a closed contour
∣∫Cf(z) dz∣≤max⁡z∈C∣f(z)∣⋅L(C)\left|\int_C f(z)\,dz\right| \le \max_{z\in C}|f(z)| \cdot L(C)
ML inequality (L(C) = length of C)

Worked Examples

  1. Parametrize the unit circle as z(t) = e^{it}, t from 0 to 2π, so dz = i e^{it} dt.

    z(t)=eit,dz=ieit dtz(t) = e^{it}, \qquad dz = ie^{it}\,dt

  2. Substitute into the integral.

    ∮Cdzz=∫02πieiteit dt=∫02πi dt\oint_C \frac{dz}{z} = \int_0^{2\pi} \frac{ie^{it}}{e^{it}}\,dt = \int_0^{2\pi} i\,dt

  3. Evaluate the elementary integral.

    ∫02πi dt=i⋅2π=2πi\int_0^{2\pi} i\,dt = i\cdot 2\pi = 2\pi i

Answer: ∮_C dz/z = 2πi (this is the seed computation behind the residue theorem).

Practice Problems

Difficulty 4/10

Compute ∮_C dz/z where C is the circle of radius 2 centered at the origin, traversed counterclockwise.

Difficulty 5/10

Compute ∫_C z² dz where C is the straight segment from z=0 to z=1 (real axis).

Difficulty 6/10

Use the ML inequality to bound |∮_C dz/z²| where C is the circle |z|=2 traversed once counterclockwise.

Quiz

A contour integral ∫_C f(z) dz is defined by:
∮_C dz/z for C the unit circle counterclockwise equals:
The ML inequality bounds a contour integral by:

Summary

  • A contour integral ∫_C f(z) dz is computed by parametrizing C as z(t) and integrating f(z(t))z'(t) over the parameter interval.
  • Contour integrals generally depend on the path, not just endpoints, unless f is analytic throughout a simply connected region containing all paths considered.
  • ∮_C dz/z = 2πi for any counterclockwise simple closed curve C enclosing the origin once — the seed identity behind the residue theorem.

References

  1. WebsiteWikipedia — Contour integration