Contour Integrals


For \(x\in\mathbb{R}, \int_a^b dxf(x) = \lim_{\Delta x\to 0}\sum f(x_i)\Delta x_i\), but \(z\) can go more places!

Complex Extension

In the complex plane, multiple paths from \(a\) to \(b\), i.e. multiple contours. Common symbol for a contour is \(\gamma\) and also has an orientation, i.e. direction, associated with it (i.e. from \(a\) to \(b\)).

The integral then is defined as expected, \(\int_\gamma dz f(z) = \lim_{n\to\infty}\sum_k^n f(z_k)(z_k-z_{k-1})\).

From this, it should be obvious that, \(0\leq |\int_\gamma dzf(z)| \leq L\gamma\cdot\sup_{z\in\gamma} f(z)\), where $|γ| = Lγ = $ length of \(\gamma\).

Darboux Inequality

This is useful for provimg different things or estimating, \(|\int_\gamma dzf(z)| \leq L\gamma\cdot\sup_{z\in\gamma} f(z)\)


We can parameterize \(\gamma\) by \(\{z(t):0\leq t\leq 1\}\), \(dz = \frac{dz}{dt}dt\). Then \(L\gamma = \int_0^1dt |\dot{z}| = \int_\gamma |dz|\)


All of the usual properties:

  • Linear
  • Integration by Parts (Undoing product rule)
  • Piecewise Integration: \(\int_\gamma dzf(z) = \int_{\gamma_1}dzf(z) + \int_{\gamma_2}dzf(z)\), where \(\gamma = \gamma_1 \# \gamma_2\). The hash denotes the concatenation of the two contours.
  • Reversing orientation brings out a negative sign: \(\int_\gamma dzf(z) = \int_{\overline{\gamma}}dzf(z)\). \(\overline{\gamma}\) and \(-\gamma\) typically refer to the same thing.
  • Antiderivatives


\(\int_\gamma dz = \lim_{n\to\infty}\sum_k^n (1)\Delta z_k = (z_1-z_0) + (z_2 - z_1) \cdots = z_n - z_0 = b-a\).

Suppose \(F(z)\) is an antiderivative of a function \(f(z)\) is holomorphic in some domain \(\mathcal{D}\). If \(f(z) = \frac{dF}{dz}(z); z\in\mathcal{D}\) then for a \(\gamma\) that remains in \(\mathcal{D}\), \(\int_\gamma dz f(z) = \int_{t_0}^{t_1} dt \frac{dz}{dt}(t)f(z(t)) = \int_{t_0}^{t_1} dt \frac{dz}{dt}(t)\frac{dF}{dz}(z(t)) = \int_{t_0}^{t_1}dt\frac{dF}{dt}(z(t)) = F(z(t_1)) - F(z(t_0)) = F(b) - F(z)\).

So for functions that are holomorphic in a region \(\mathcal{D}\), we have the Fundamental Theorem of Calculus you would expect. Because derivatives are what we expect, c.f. real counterparts, so are (indefinite) integrals. E.g. \(\int dz z^n = \frac{1}{n+1}z^{n+1}\), \(\int dz\cos z = \sin z\), etc.

The indefinite integrals arise from letting the endpoint be arbitrary, i.e. let the integral be a function of \(z\).

An important implication is that, if \(f(z)\) is holomorphic and the integral only depends on the end points then in the region \(\mathcal{D}\) the integral is path independent. Further, because of path independence in this region, we have that a closed contour yields zero in this region \(\mathcal{D}\), where the function is holomorphic. I.e. the region \(\mathcal{D}\) needs to be simply connected to allow path deformation.


\(z = x+iy = r\exp(i\theta)\) \(f(z) = u + iv\)
\(z(t) = x(t)+iy(t) = r(t)\exp(i\theta(t))\) \(f(z) = f(z(t)) = u(t) + iv(t)\)

Choose whichever represenation is suited to you path \(\gamma\). \(dz = dx + idy = \dot{z}dt = \left(\frac{dx}{dt} + i\frac{dy}{dt}\right)dt = d(r\exp(i\theta)) = ir\exp(i\theta)d\theta + \exp(i\theta) dr\)

Example 1

Positively oriented semicircle with the line on the real axis and the arc curving to the positive imaginary axis (counter clockwise). \(z=r\exp(i\theta),dz=ir\exp(i\theta)d\theta+dr\exp(i\theta)\). \(\int_\gamma dz z^2 = \int_{\gamma_1} dz z^2 + \int_{\gamma_2}dzz^2 = \int_{-r}^rdx (x+i0)^2 + \int_0^\pi (ir\exp(i\theta)d\theta) (r\exp(i\theta))^2 = \int_{-r}^rdx x^2 + r^3\int_0^\pi i\text{d}\theta\exp(i3\theta)=\frac{r^3}{3}-\frac{(-r)^3}{3} + r^3\left[\frac{\exp(3i\pi)}{3}-\frac{\exp(0)}{3}\right]=\frac{2r^3}{3}-\frac{2r^3}{3}=0\).

Alternatively for one of the integrals: \(\int_{-R}^R dr\exp(i0)r^2\exp(2i0)=\exp(3i0)\frac{2r^3}{3}\).

Example 2 - Parameterization

\(f(z)=\exp(z)\). \(x(t)=a t\) \(y(t)=b t\) \(z=at+bti\) \(dz=(a+bi)dt\) \(\exp(x(t)+iy(t)) = \exp(x(t)+iy(t))=\exp(t(a+bi))\) \(\int_\gamma dz f(z) = \int_0^adx f(x+iy) + \int_0^b (idy) f(x+iy) + \int_1^0 [(a+bi)dt] f(x(t)+iy(t)) = \int_0^a dx f(x+i0) + \int_0^b dy f(a+iy) + \int_1^0 dt f(at+ibt) = (\exp(a)-\exp(0)) + \frac{i}{i}(\exp(a+bi)-\exp(a)) + \frac{a+bi}{a+bi}(\exp(0)-\exp(a+bi)) = 0\)

Example 3 - Singularities


\(\int_\gamma dz f(z) = \int_0^{2\pi}izd\theta f(z) = i\int_0^{2\pi}d\theta = 2\pi i\)

Author: Christian Cunningham

Created: 2023-06-25 Sun 02:24