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:


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.


\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.


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