Contour Integrals

Complex integration extends the notion of integration to functions of a complex variable along curves in the complex plane. The key distinction from real integration is that the path of integration matters fundamentally.

Curves in the Complex Plane

Definition3.1Smooth curve

A smooth curve (or smooth arc) in $\mathbb{C}$ is a continuous function $\gamma: [a,b] \to \mathbb{C}$ such that $\gamma'(t)$ exists, is continuous, and satisfies $\gamma'(t) \neq 0$ for all $t \in [a,b]$ . A piecewise smooth curve (or contour) is a finite concatenation of smooth curves.

Definition3.2Simple and closed curves

A curve $\gamma: [a,b] \to \mathbb{C}$ is closed if $\gamma(a) = \gamma(b)$ . It is simple if $\gamma(t_1) \neq \gamma(t_2)$ for all $t_1 \neq t_2$ except possibly $\gamma(a) = \gamma(b)$ . A simple closed curve is called a Jordan curve.

ExampleStandard contours

The circle $\gamma(t) = z_0 + re^{it}$ , $t \in [0, 2\pi]$ , traverses the circle $|z - z_0| = r$ counterclockwise.
The line segment from $z_1$ to $z_2$ : $\gamma(t) = (1-t)z_1 + tz_2$ , $t \in [0,1]$ .
The rectangular contour with vertices $\pm R \pm iR$ is a piecewise smooth closed curve.

The Complex Line Integral

Definition3.3Complex line integral

Let $f$ be continuous on a smooth curve $\gamma: [a,b] \to \mathbb{C}$ . The complex line integral (or contour integral) of $f$ along $\gamma$ is

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

For a piecewise smooth curve $\gamma = \gamma_1 + \gamma_2 + \cdots + \gamma_n$ , we define $\int_\gamma f\,dz = \sum_{k=1}^n \int_{\gamma_k} f\,dz$ .

RemarkPath dependence

Unlike definite integrals in $\mathbb{R}$ , the value of $\int_\gamma f(z)\,dz$ generally depends on the choice of path $\gamma$ , not just on the endpoints. This path dependence is eliminated precisely when $f$ has an antiderivative, which connects to the Cauchy--Goursat theorem.

ExampleIntegral of $z^2$

Let $\gamma(t) = t + it$ , $t \in [0,1]$ (the line from $0$ to $1+i$ ). Then $\gamma'(t) = 1+i$ and

$\int_\gamma z^2\,dz = \int_0^1 (t+it)^2(1+i)\,dt = (1+i)^3 \int_0^1 t^2\,dt = (1+i)^3 \cdot \frac{1}{3} = \frac{-2+2i}{3}.$

Since $z^2$ has antiderivative $z^3/3$ , we verify: $\frac{(1+i)^3}{3} - 0 = \frac{-2+2i}{3}$ .

The ML Inequality

Theorem3.1ML inequality

If $f$ is continuous on a contour $\gamma$ of length $L$ and $|f(z)| \leq M$ for all $z$ on $\gamma$ , then

$\left|\int_\gamma f(z)\,dz\right| \leq ML.$

RemarkEstimation applications

The ML inequality is one of the most frequently used tools in complex analysis. It is essential for:

Showing that integrals over certain contours vanish as the contour grows (Jordan's lemma).
Establishing convergence of integral representations.
Bounding error terms in asymptotic expansions.

ExampleBounding an integral

Let $\gamma$ be the upper semicircle $|z| = R$ , $\text{Im}(z) \geq 0$ . For $f(z) = 1/(z^2+1)$ with $R > 1$ :

$|f(z)| = \frac{1}{|z^2+1|} \leq \frac{1}{R^2 - 1}, \quad L(\gamma) = \pi R.$

Therefore $\left|\int_\gamma \frac{dz}{z^2+1}\right| \leq \frac{\pi R}{R^2 - 1} \to 0$ as $R \to \infty$ .

Fundamental Theorem of Contour Integration

Theorem3.2Fundamental theorem of contour integration

Let $D \subseteq \mathbb{C}$ be a domain. If $f: D \to \mathbb{C}$ is continuous and has an antiderivative $F$ on $D$ (i.e., $F'(z) = f(z)$ for all $z \in D$ ), then for any contour $\gamma$ in $D$ from $z_1$ to $z_2$ :

$\int_\gamma f(z)\,dz = F(z_2) - F(z_1).$

In particular, $\oint_\gamma f(z)\,dz = 0$ for every closed contour $\gamma$ in $D$ .

ExampleUsing antiderivatives

Since $e^z$ is its own antiderivative on all of $\mathbb{C}$ , for any contour $\gamma$ from $0$ to $\pi i$ :

$\int_\gamma e^z\,dz = e^{\pi i} - e^0 = -1 - 1 = -2.$

However, $1/z$ does not have a single-valued antiderivative on $\mathbb{C} \setminus \{0\}$ , so $\int_\gamma dz/z$ depends on the path.

The Integral of $1/z$

RemarkThe key integral

The integral $\oint_{|z|=1} \frac{dz}{z} = 2\pi i$ is the most important computation in complex analysis. More generally, for any circle $|z - z_0| = r$ traversed counterclockwise:

$\oint_{|z-z_0|=r} \frac{dz}{z - z_0} = 2\pi i, \qquad \oint_{|z-z_0|=r} (z-z_0)^n\,dz = 0 \quad (n \neq -1).$