Evaluating a complex contour I need to show the following result:
$$
\int_{-\infty}^\infty \frac{1}{(1+x^2)^{n+1}}dx\, = \frac{1\cdot 3\cdot\ldots\cdot(2n-1)}{2\cdot 4\cdot\ldots\cdot(2n)}\pi 
$$
With n=1,2,3,...
This function has a pole at i and -i.
I've tried a semicircle in the upperhalf of the plain, but the residue then goes to infinity. I've also tried a rectangle in the upperhalf that stays beneath i, but all 3 sides that do not include the integral we're looking for go to zero because of the R in the denominator.
Anyone with tips?

Integrating over a large semicircle in the upper halfplane works well.
Write $$\frac{1}{(1+z^2)^{n+1}} = \dfrac{\dfrac{1}{(z+i)^{n+1}}}{\quad(z-i)^{n+1}\quad}.$$
Hence, the residue at $z=i$ is the $1/n!$ times the $n$th derivative of $1/(z+i)^{n+1}$ evaluated at $z=i$, i.e.
$$
\begin{split}
\operatorname{Res}\limits_{z=i} \frac{1}{(1+z^2)^{n+1}} &= 
\frac{1}{n!}\frac{d^n}{dz^n}\left( \frac{1}{(z+i)^{n+1}} \right)\Bigg|_{z=i} \\
&= \frac{1}{n!}(-1)^n(n+1)(n+2)\cdots(2n)\frac{1}{(2i)^{2n+1}} \\
&=\frac{(n+1)(n+2)\cdots(2n)}{n!\cdot2^{2n}\cdot 2i} \\
&= \frac{n!\cdot(n+1)\cdots(2n)}{(n!)^2\cdot 2^{2n}\cdot 2i} \\
&= \frac{(2n)!}{(2\cdot 4\cdot 6\cdots(2n))^2\cdot 2i} \\
&= \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot 4\cdot 6\cdots(2n)\cdot 2i} \\
\end{split}
$$
I think you can do the last step yourself.
