Please, help to evaluate the integral (using residue)
$\displaystyle \int_{0}^{\infty} \frac{1}{(x^{2}+1)^n}dx$,
n - natural.
Thx.
Oh, a chance to do a little CA. Let's see. I will do the case -inf..inf and divide by 2.
$\displaystyle \int_{-\infty}^{\infty}\frac{1}{(x^{2}+1)^{n}}dx$
There is a pole of order n at z=i. The residue is
$\displaystyle \frac{1}{(n-1)!}\left[\frac{d^{n-1}}{dz^{2}}\left(\frac{1}{(z+i)^{n}}\right)\right]$
$\displaystyle =\frac{1}{(n-1)!}\left[\frac{(-n)(-n-1)......(-n-n+2)}{(z+i)^{2n-1}}\right]$
$\displaystyle =\frac{(-1)^{n-1}n(n+1)....(2n-2)(-1)^{n}i}{n^{2n-1}(n-1)!}$
Almost there, Can you take over?.