# Residue calculus

August 10, 2012 Leave a comment

There’s a neat trick we can use to integrate real integrals using complex analysis. These can be made arbitrarily complicated, but I’ll give you a simple example. Compute the integral:

This is another way to write down

Let’s change all the s to s, and pretend we’re integrating along the path in the complex plane. Our notion of integrating in is defined in such a way that this makes sense. Okay, but we normally integrate loops, not paths, so let’s complete a full loop is the semi-circle in the upper half-plane with base and radius . Then we can break the path into two pieces:

The first piece is the integral we want to compute, and the second is the curved part of the semi-circle. Together they make a closed loop whose integral we can actually calculate using the residue theorem. Since our function is meromorphic on all of , we need to simply figure out which poles are inside the semicircle , and find their residues. We can tell that the only poles are at and , of which only is in the semicircle (for every ). It’s residue we compute as .

Thus, the left hand side, via the residue theorem, must be .

Now we just need to show that the semi-circular arc’s contiribution is neglibible for large . Then, taking yields

Indeed,

which decreases to zero as .

I’m awful at integration in general, so I’m not sure if this integral can be done using integration by parts, or a tricky -substitution. There are definitely some integrals though, for which those standard methods just don’t cut it.