Contour Integrals

Contour integration is the backbone of evaluating real definite integrals that resist elementary methods — physicists and engineers routinely compute integrals like $\int_{-\infty}^\infty \frac{\cos x}{x^2+1}dx$ by closing a contour in the complex plane and invoking the machinery this article sets up. In electrical engineering and signal processing, contour integrals underlie the inverse Laplace and Z-transforms, used to reconstruct time-domain signals from frequency-domain representations. In physics, contour integration appears throughout quantum field theory (Feynman propagators, dispersion relations) and in statistical mechanics (partition function analysis via saddle-point and steepest-descent contours). In pure mathematics, contour integrals are the essential ingredient in proving the Fundamental Theorem of Algebra, in defining and computing residues, and in constructing analytic continuations of functions like the Riemann zeta function. Cryptography and coding theory occasionally invoke contour-integral representations when analyzing generating functions combinatorially (e.g., extracting coefficients via Cauchy's formula). Anywhere a problem can be recast as "integrate an analytic function around a loop," this article's definitions are the entry point.

:::quiz[If $\gamma(t) = 5e^{it}$ for $t \in [0, 2\pi]$, what is $\int_\gamma \frac{1}{z}\,dz$?]

• $0$
• $\pi i$
• $2\pi i$ ✓
• $10\pi i$

:::quiz[For the contour $\gamma(t) = 2 + 3e^{it}$, $t\in[0,2\pi]$, what is $\int_\gamma (z-2)^{-2}\,dz$?]

• $2\pi i$
• $0$ ✓
• $-2\pi i$
• $6\pi i$

:::quiz[Why does $\int_\gamma \frac{dz}{z-1}$ fail to be defined for $\gamma(t)=e^{it}$, $t\in[0,2\pi]$?]

• Because $\gamma$ is not piecewise-smooth
• Because the integrand is discontinuous at a point lying on the trace of $\gamma$ ✓
• Because $\gamma$ is not a closed contour
• Because $1/(z-1)$ is not linear

Ask the tutor: How does the ML-inequality get used to prove that certain contour integrals vanish in the limit as a radius grows or shrinks to zero?

Think of $\gamma(t)$ as tracing out the position of a moving particle in the plane at time $t$, and $\gamma'(t)$ as its instantaneous velocity vector (a complex number encoding both speed and heading). The quantity $f(\gamma(t))\,\gamma'(t)\,dt$ is a complex number for each instant: it multiplies the "local value" of $f$ felt by the particle by the tiny displacement $\gamma'(t)\,dt$ the particle makes in that instant. Summing (integrating) these complex products over the whole time interval accumulates a single complex number — the contour integral. Crucially, this is not the same as integrating $f$ against arc length $|dz|$ (that would be a different, real-valued-weighted object used to define length or to prove the ML-inequality below). The contour integral retains directional information: the complex factor $\gamma'(t)$ rotates and scales the value $f(\gamma(t))$ according to the direction the curve is heading at that moment. This is why reversing the direction of travel flips the sign of the integral, and why re-tracing the same geometric curve at different speeds (but same direction) gives the same answer — speed is a red herring, direction is what matters. A second useful mental picture: writing $f = u + iv$ and $\gamma(t) = x(t) + iy(t)$, expanding the product $f(\gamma(t))\gamma'(t)\,dt$ shows that a contour integral packages two real line integrals (of the sort seen in vector calculus) into one complex number:

This connects contour integration to the line integrals of vector fields you may have seen for computing work and flux, but bundled with the special complex-multiplicative structure that will later produce Cauchy's theorem.

:::svg[The contour $\gamma(t) = e^{it}$ for $t\in[0,\pi]$: velocity $\gamma'(t) = ie^{it}$ is tangent, rotated $90°$ ahead]

Example 1: Integrating $z^n$ around a circle not centered at the origin. Let $\gamma(t) = 2 + 3e^{it}$ for $t \in [0, 2\pi]$ — the circle of radius $3$ centered at $z_0 = 2$, traversed once counterclockwise. We compute $\int_\gamma (z-2)^n\,dz$ for an integer $n$ (positive, negative, or zero), and separately note what happens for a shifted power like $z^n$ itself, to see the role of centering. Since $\gamma(t) - 2 = 3e^{it}$, we have $\gamma'(t) = 3ie^{it}$, so

Case $n \neq -1$: since $n+1 \neq 0$,

because $e^{i(n+1)2\pi} = 1$ for any integer $n+1$. So $\int_\gamma (z-2)^n\,dz = 0$ for every integer $n \ne -1$ (including negative $n$, for which $(z-2)^n$ is a well-defined continuous function on the trace, since $z_0=2$ is never on the circle $|z-2|=3$... wait, actually $z=2$ is the center, not on the trace, so $(z-2)^n$ for $n<0$ is perfectly continuous on the circle $|z-2|=3$). Case $n = -1$: here $n+1 = 0$ and $e^{i\cdot 0 \cdot t} = 1$ identically, so

giving

So we have derived the fundamental fact

independent of the radius $3$ — a first hint of the deep rigidity that Cauchy's theorem will explain. Note this computation required re-centering at $z_0=2$; if instead we integrated $z^{-1}$ (not $(z-2)^{-1}$) over this same circle, the singularity at $z=0$ lies inside the disk bounded by $|z-2|=3$ (since $|0-2|=2<3$) and a direct substitution $z = 2+3e^{it}$ gives a messier integral not reducible to a single clean exponential — illustrating why choosing the parametrization to match the geometry of the singularity matters. We do not evaluate that harder integral here; it becomes transparent once the residue and Cauchy machinery (foreshadowed by this article) is available.

Example 2: A non-closed contour and the ML-inequality in action. Let $\gamma(t) = t + it^2$ for $t \in [0,1]$ — a parabolic arc from $\gamma(0) = 0$ to $\gamma(1) = 1+i$. We compute $\int_\gamma z\,dz$ two ways and also bound it via ML. Direct computation. Here $\gamma'(t) = 1 + 2it$, so

Expand the integrand:

So

Check via antiderivative. Since $z \mapsto z^2/2$ is an antiderivative of $z\mapsto z$ on all of $\mathbb{C}$, and this is a smooth path, the Fundamental-Theorem-of-Calculus-style formula (a fact that follows from the chain rule exactly as in real calculus, and which will be developed fully as the Fundamental Theorem for Contour Integrals) gives

matching our direct computation exactly, and confirming this integral depends only on endpoints here (a special feature of $f(z)=z$ having an antiderivative, not a general phenomenon for arbitrary contours and functions). ML bound. Let's sanity-check with the ML-inequality (Part 5 of the theorem). On $[0,1]$, $|\gamma(t)| = |t+it^2| = t\sqrt{1+t^2} \le \sqrt{2}$, so $M = \sup_{t\in[0,1]}|f(\gamma(t))| = \sqrt 2$ works as a bound for $f(z)=z$ on the trace. The length is

Numerically, $\int_0^1 \sqrt{1+4t^2}\,dt \approx 1.4789$ (this can be found via the standard antiderivative $\frac{t}{2}\sqrt{1+4t^2} + \frac14 \sinh^{-1}(2t)$, evaluating to $\frac12\sqrt5 + \frac14\ln(2+\sqrt5) \approx 1.1180+0.3609 \approx 1.4789$). So the ML bound predicts

while the true value has $|i| = 1$. The bound $1 \le 2.091$ holds comfortably, as it must — the inequality is an upper bound, not an exact formula, and this example shows it can be quite loose; its value lies in giving guaranteed control without needing an exact antiderivative, which is essential for functions where no elementary antiderivative exists.