Residue theorem

Statement

Let $U \subseteq \mathbb{C}$ be a simply connected open set. This means that $U$ is connected, and each closed path in $U$ can be continuously deformed in $U$ to a constant path. Let $a_1, \dots, a_n$ be distinct points of $U$, and let $f$ be holomorphic on $U \setminus \{a_1, \dots, a_n\}$. Let $\gamma$ be a closed, continuous and piecewise continuously differentiable path in $U \setminus \{a_1, \dots, a_n\}$. Then $$\oint_\gamma f(z) \, dz = 2 \pi i \sum_{k=1}^{n} n(\gamma, a_k) \operatorname{Res}(f, a_k),$$ where $n(\gamma, a_k)$ is the winding number of $\gamma$ around $a_k$ and $\operatorname{Res}(f, a_k)$ is the residue of $f$ at $a_k$.

Remarks

If $\gamma$ is a simple closed curve traveled once counterclockwise, then $\oint_\gamma f(z) \, dz$ is $2 \pi i$ times the sum of the residues at the points $a_k$ inside $\gamma$. With no points $a_k$, the theorem is Cauchy's integral theorem. For example, $\oint_{|z| = 2} \frac{dz}{z^2 + 1} = 2 \pi i \left( \frac{1}{2i} - \frac{1}{2i} \right) = 0$, by the residues at the simple poles $i$ and $-i$. The theorem is the main tool for the evaluation of real integrals.

Explanations

L1 human

To integrate around a closed curve, you only need to know the residues at the bad points inside the curve. Each bad point contributes $2 \pi i$ times its residue, once for each turn of the curve around it.