TITLE: Applications of the Residue Theorem to the Evaluation of Integrals and Sums
QUESTION [5 upvotes]: Evaluate the integral 

$$\int_{-\infty}^{\infty} \frac{1}{(1 + x^2)^{n+1}} dx. $$ 

I know that it equals $2\pi i$(the sum of the residues; at $z_k$) where $z_k$ are the poles of the function. I can evaluate this without the $n+1$ but that is throwing me for a serious loop.

REPLY [7 votes]: For this case, consider
$$\oint_C \frac{dz}{(1+z^2)^{n+1}} $$
where $C$ is a semicircle of radius $R$ in the upper half plane.  By the residue theorem, the contour integral is equal to $i 2 \pi$ time the residue at the pole $z=i$.  Noting that the integral about the semicircle vanishes as $1/R^{2 n+1}$ as $R \to \infty$, we have
$$\int_{-\infty}^{\infty} \frac{dx}{(1+x^2)^{n+1}}  = i 2 \pi \operatorname*{Res}_{z=i} \frac{1}{(1+z^2)^{n+1}}$$
Now,
$$\operatorname*{Res}_{z=i} \frac{1}{(1+z^2)^{n+1}} = \frac{1}{n!} \left [\frac{d^n}{dz^n} \frac{1}{(z+i)^{n+1}}\right ]_{z=i} = \frac{(2 n)!}{(n!)^2} \frac{(-1)^n}{(2 i)^{2 n+1}}$$
Therefore
$$\int_{-\infty}^{\infty} \frac{dx}{(1+x^2)^{n+1}}  = \frac{\pi}{2^{2 n}} \binom{2 n}{n}$$