# Residue calculus

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:

$\displaystyle \displaystyle\int_{-\infty}^\infty\frac1{(1+x^2)^3}dx$

This is another way to write down

$\displaystyle \displaystyle\lim_{a\rightarrow\infty}\int_{-a}^a\frac1{(1+x^2)^2}dx.$

Let’s change all the ${x}$s to ${z}$s, and pretend we’re integrating along the path ${[-a,a]}$ in the complex plane. Our notion of integrating in ${{\mathbb C}}$ 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 ${\gamma_a}$ is the semi-circle in the upper half-plane with base ${[-a,a]}$ and radius ${a}$. Then we can break the path into two pieces:

$\displaystyle\int_{\gamma_a}\frac1{(1+z^2)^{2}}dz=\int_{-a}^a\frac1{(1+x^2)^{2}}dx+\int_0^\pi\frac1{(1+(ae^{i\theta})^2)^{2}}d\theta.$

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 ${{\mathbb C}}$, we need to simply figure out which poles are inside the semicircle ${\gamma_a}$, and find their residues. We can tell that the only poles are at ${i}$ and ${-i}$, of which only ${i}$ is in the semicircle ${\gamma_a}$ (for every ${a>1}$). It’s residue we compute as $-i/4$.

Thus, the left hand side, via the residue theorem, must be ${2\pi i\cdot (-i/4)=\pi/2}$.

Now we just need to show that the semi-circular arc’s contiribution is neglibible for large ${a}$. Then, taking ${a\rightarrow\infty}$ yields

$\displaystyle \displaystyle\lim_{a\rightarrow\infty}\int_{-a}^a\frac1{(1+x^2)^2}dx=\pi/2.$

Indeed,

$\displaystyle \begin{array}{rcl} \left|\displaystyle\int_0^\pi\frac1{(1+(ae^{i\theta})^2)^{2}}d\theta\right| &\le& \displaystyle\int_0^\pi\left|\frac1{(1+(ae^{i\theta})^2)^{2}}\right|d\theta\\ &=& \displaystyle\int_0^\pi\frac1{(1+a^2)^{2}}d\theta\\ &=& \frac{\pi}{(1+a^2)^2} \end{array}$

which decreases to zero as ${a\rightarrow\infty}$.

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 ${u}$-substitution. There are definitely some integrals though, for which those standard methods just don’t cut it.