How to choose the right contour in complex integration?

Question

I'm having some conceptual difficulty. In evaluating integrals via contour integration, the choice of the right contour seems rather like trial-and-error. For example in the integration of cosh(ax)/cos(x) over 0 to infinity under the condition mod.(a)<1 I have to assume a rectangular contour. Why can I not take a circular contour instead?

Answer 1

There isn't always a systematic method for how to choose your contour. In fact, in areas that make heavy use of tools like the method of steepest descent, you often run into quite complex contour integrals where the main problem is to figure out a contour that works.

That said, there is some rhyme and reason to how one might build up a useful contour, which is based on standard contours used for relatively simple contour integrals. To give some examples:

1. Rectangular contours are often used for dealing with integrands that are effectively Fourier transforms (involving terms like $e^{i\theta x}$ in the numerator, sometimes implicitly through terms like $\cos(\theta x)$, because $e^{i\theta z}$ transforms very predictably when you go from $z=x$ to $z = x+iy$. For example this is one commonly used method to show that the Fourier transform of a Gaussian is another Gaussian. Putting semicircular dents in the contour to avoid poles on the real axis should be relatively self-explanatory for these integrals.

2. Keyhole contours are often encountered with integrands that are implicitly Laplace transforms, such as those that involve $x^\alpha = e^{\alpha\log x}$. Because of the properties of the complex logarithm, these integrands exhibit a phase shift when you go from the contour $z = t + i \epsilon$ to $t - i\epsilon$ rather than the integrals along those contours cancelling.

3. Simple semicicular contours often work well with rational functions since their decay as $z\to\infty$ is predictable. That really is the main use case for semicircular contours, when all the terms involved decay predictably at infinity.

4. Integrals involving terms like $\sin(x)$ or $\cos(x)$ in the denominator sometimes benefit from integrating over arcs of the unit circle, because they may simplify to rational functions of $z = e^{ix}$. Sometimes this effectively recovers substitution methods like the tangent half-angle substitution.

So in summary, there is a bit of a taxonomy for the simple cases of contour integration, and for more complex integrands you might need to combine ideas from the simple cases. Kind of like how regular Riemann integration works.