Residue Theorem

Contour integrals as 2π i times the sum of enclosed residues.

The idea

Theorem (Residue Theorem).

Let $\gamma$ be a positively oriented simple closed contour, and let $f$ be holomorphic on $\gamma$ and inside it except at finitely many isolated singularities $z_{1}, \ldots, z_{k}$ lying strictly inside. Then $\oint_{\gamma} f(z)\,dz = 2\pi i \sum_{j=1}^{k} \operatorname{Res}_{z = z_{j}} f,$ where the residue $\operatorname{Res}_{z = z_{j}} f$ is the coefficient $a_{-1}$ of $(z - z_{j})^{-1}$ in the Laurent series of $f$ about $z_{j}$.

The coefficient $a_{-1}$ alone decides the integral because every other term integrates to zero. Expand $f$ about a singularity and integrate term by term around a small circle centred there. For $n \neq -1$, the term $(z - z_{0})^{n}$ has the antiderivative $(z - z_{0})^{n+1}/(n+1)$, a single-valued function on the circle, so its integral around the loop vanishes. For $n = -1$ the antiderivative would have to be a logarithm, and no logarithm is continuous all the way around a loop enclosing $z_{0}$, since the argument advances by $2\pi$ in one circuit; that term contributes $2\pi i$ times its coefficient.

Finally, $f$ is holomorphic in the region between $\gamma$ and small circles about the singularities, so $\gamma$ deforms onto those circles without changing the integral, and the contributions add. A singularity lying outside $\gamma$ is never encircled and contributes nothing.

Ways to work on it

• Walkthrough. Residues, the simple-pole formula, and which poles are enclosed.
• Practice. Integrate a one-pole function around a circle.
• Hardest. Sum two residues and assemble the contour integral.

Not sure where to start? Take the ten-question placement test.