Contour Integration

Contour integration is a method of evaluating integrals of functions along oriented curves in the complex plane. It is an extension of the usual integral of a function along an interval in the real number line. Contour integrals may be evaluated using direct calculations, the Cauchy integral formula, or the residue theorem.

A curve in the complex plane is a set of points parameterized by some function $[a,b] \to \mathbb{C}$, usually denoted $\gamma$ parameterized by $z$. $\gamma$ is differentiable if $z'(t)$ exists for all $t \in [a,b]$.

Let $f: U \to \mathbb{C}$ and $\gamma$ be a differentiable curve parameterized by $z: [a,b] \to \mathbb{C}$. The integral of $f(z) \ dz$ along $\gamma$ is defined as

$$\int_\gamma f(z) \ dz = \int_a^b f\big(z(t)\big)z'(t) \ dt.$$

$($If $f: U \to \mathbb{C}$ is a holomorphic function, then one may think of it as a function from a subset of $\mathbb{R}^2$ to the whole $\mathbb{R}^2$. In other words, $f$ implicitly defines a vector field on $\mathbb{R}^2$. Accordingly, one naturally integrates holomorphic functions in the same way as one integrates vector fields, using line integrals.$)$

Some contour integrals may be computed by hand. For example:

Let $\gamma$ denote the circle centered at $O$ with radius $1$. Then compute

$$\int_{\gamma} \frac{1}{z} \, dz.$$

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First, parameterize the circle by $\gamma(t) = e^{2\pi i t}$, so that as $t$ ranges over $[0,1]$, the entire circle is traced out once. Note that $\gamma(0) = \gamma(1)$; a curve with this property is called closed.

With this parameterization, one computes

$$\int_{\gamma} \frac{1}{z} \, dz = \int_{0}^{1} \frac{1}{\gamma(t)} \gamma'(t) \, dt = \int_{0}^{1} e^{-2\pi i t} \cdot 2\pi i e^{2\pi i t} \, dt = \int_{0}^{1} 2\pi i \, dt = 2\pi i.\ _\square$$

Note that in the above example, the function $\frac{1}{z}$ is not holomorphic on the region interior to $\gamma$, since it is not defined at $z=0$.

Take $f: U \to \mathbb{C}$ to be holomorphic. The primitive of $f$ is a function $F: U \to \mathbb{C}$ such that $F'(z) = f(z)$ for all $z \in U$.

Suppose $\gamma$ is a curve parametrised by $z: [a,b] \to \mathbb{C}$ and $f$ has a primitive. Then, since complex differentiation follows the chain rule,

$$\begin{aligned} \int_\gamma f(z) \ dz &= \int_a^b f\big(z(t)\big) z'(t) \ dt \\ &= \int_a^b F'\big(z(t)\big) z'(t) \ dt \\ &= \int_a^b \frac{d}{dt} F\big(z(t)\big) \ dt \\ &= F\big(z(b)\big)-F\big(z(a)\big). \end{aligned}$$

In particular, if $\gamma$ is a closed curve, i.e. $z(b) = z(a)$, then $\displaystyle \int_\gamma f(z) \ dz = 0$. There is a result that all holomorphic functions have primitives. Thus, the only nontrivial contour integrals along closed curves are those which enclose poles of the function being integrated.