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Contour Integration — Definition, Formula & Examples

Contour integration is the process of integrating a complex-valued function along a directed curve (called a contour) in the complex plane. It generalizes ordinary integration to paths that aren't just segments of the real number line.

Given a complex-valued function f(z)f(z) and a piecewise-smooth curve γ:[a,b]→C\gamma:[a,b]\to\mathbb{C}, the contour integral of ff along γ\gamma is defined as ∫γf(z) dz=∫abf(γ(t)) γ′(t) dt\int_\gamma f(z)\,dz = \int_a^b f(\gamma(t))\,\gamma'(t)\,dt, where γ′(t)\gamma'(t) is the derivative of the parametrization.

Key Formula

∫γf(z) dz=∫abf(γ(t)) γ′(t) dt\int_\gamma f(z)\,dz = \int_a^b f(\gamma(t))\,\gamma'(t)\,dt
Where:
  • f(z)f(z) = Complex-valued function being integrated
  • γ(t)\gamma(t) = Parametrization of the contour, mapping $[a,b]$ into $\mathbb{C}$
  • γ′(t)\gamma'(t) = Derivative of the parametrization with respect to $t$
  • [a,b][a,b] = Interval of the real parameter $t$

How It Works

To evaluate a contour integral, you parametrize the curve γ\gamma using a real variable tt, substitute z=γ(t)z = \gamma(t) and dz=γ′(t) dtdz = \gamma'(t)\,dt, then compute the resulting ordinary integral over tt. For many closed contours, the Cauchy integral theorem states the integral is zero when ff is analytic inside and on the contour. When ff has singularities enclosed by the contour, the residue theorem lets you evaluate the integral by summing residues, often avoiding direct parametrization entirely.

Worked Example

Problem: Evaluate ∫γ1z dz\int_\gamma \frac{1}{z}\,dz where γ\gamma is the unit circle traversed counterclockwise.
Parametrize the contour: The unit circle can be parametrized as γ(t)=eit\gamma(t) = e^{it} for t∈[0,2π]t \in [0, 2\pi]. Then γ′(t)=ieit\gamma'(t) = ie^{it}.
z=eit,dz=ieit dtz = e^{it},\quad dz = ie^{it}\,dt
Substitute into the integral: Replace zz and dzdz in the integrand.
∫02π1eit⋅ieit dt=∫02πi dt\int_0^{2\pi} \frac{1}{e^{it}} \cdot ie^{it}\,dt = \int_0^{2\pi} i\,dt
Evaluate: The integrand simplifies to a constant, so the integral is straightforward.
i⋅2π=2πii \cdot 2\pi = 2\pi i
Answer: ∫γ1z dz=2πi\displaystyle\int_\gamma \frac{1}{z}\,dz = 2\pi i

Why It Matters

Contour integration is central to complex analysis and appears throughout physics and engineering — for instance, in evaluating real improper integrals that resist standard calculus techniques, computing inverse Laplace transforms in signal processing, and analyzing fluid flow around obstacles.

Common Mistakes

Mistake: Forgetting to include the γ′(t)\gamma'(t) factor when substituting the parametrization.
Correction: The differential dzdz equals γ′(t) dt\gamma'(t)\,dt, not just dtdt. Always multiply the integrand by γ′(t)\gamma'(t) after substitution.