Contour Integration

Parametrize the path, then integrate along it in the complex plane.

The idea

A contour integral integrates a complex function along a path in the plane. Over a real interval, the integral $\int_{a}^{b} f(x)\,dx$ never names its route from $a$ to $b$, because on a line there is only one; in the plane many routes join two points, so the integral must say which one it follows.

The route is a contour: a path $\gamma$ described by a parametrization $\gamma(t)$ for $a \le t \le b$, where $t$ is a real parameter and $\gamma(t)$ is the point of the plane reached at time $t$. As $t$ advances by $dt$, the point moves by $dz = \gamma'(t)\,dt$, and that move contributes $f(\gamma(t))\,dz$. Adding the contributions defines $\int_{\gamma} f(z)\,dz = \int_{a}^{b} f(\gamma(t))\,\gamma'(t)\,dt,$ read as the integral of $f$ along $\gamma$. The right-hand side is an ordinary integral over a real interval whose integrand is complex, so we can evaluate it by integrating the real and imaginary parts separately, and its value does not depend on which parametrization of the path we chose.

Two standard contours recur. The segment from $z_{0}$ to $z_{1}$ is parametrized as $\gamma(t) = z_{0} + t(z_{1} - z_{0})$ for $0 \le t \le 1$; the circle of radius $R$ about the origin is $\gamma(t) = Re^{it}$ for $0 \le t \le 2\pi$, which runs once counterclockwise. A closed contour, one that returns to its start, is written $\oint_{\gamma}$.

Ways to work on it

• Walkthrough. The parametrization formula, a segment integral, and 1/z dz = 2π i.
• Practice. Integrate a power of z around the unit circle.
• Hardest. A loop integral on a larger circle, and a bound on how large it can be.

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