Evaluate $\int_{-\infty}^{\infty}\frac{1}{(x^2+4)^5}dx$ $$\int_{-\infty}^{\infty}\frac{1}{(x^2+4)^5}dx$$
I am trying to use residue.
We first need to find the singularities, $x^2+4=0\iff x=\pm 2i$
Just $2i$ is in the positive part of $i$ so we take the limit
$lim_{z\to 2i}\frac{1}{(z^2+4)^5}$ but the limit is $0$

set $x= 2t $ and then $t^2 = s$
Then 
$$ \int_{-\infty}^{\infty} \frac{dx}{(x^2 + 4 )^5}=\frac{2}{4^5}\int_{-\infty}^{\infty} \frac{dx}{(x^2 + 1 )^5}=\frac{1}{4^4}\int_{0}^{\infty} \frac{dt}{(t^2 + 1 )^5} =\frac{1}{2.4^4}\int_{0}^{\infty} \frac{s^{-1/2}ds}{(s + 1 )^5}  $$
But 
$$ \int_{0}^{\infty} \frac{s^{-1/2}ds}{(s + 1 )^5}=\int_{0}^{\infty} \frac{s^{1/2-1}ds}{(s + 1 )^{1/2+ 9/2}} = B(1/2,9/2) =\frac{\Gamma(1/2)\Gamma(9/2)}{\Gamma(5)}  $$
and remember that the beta function is given by 
$$B(x,y) =\int_{0}^{\infty} \frac{s^{x-1}ds}{(s + 1 )^{x+ y}} =\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+xy)}.  $$
also $\Gamma(1/2) = \sqrt{\pi}$, $\Gamma(9/2) = \Gamma(4+1/2)= (3+1/2)(2+1/2)(1+1/2)(1/2)\Gamma(1/2) = \Gamma(4+1/2)= (3+1/2)(2+1/2)(1+1/2)(1/2)\sqrt{\pi}$ and $\Gamma(5) = 4!$
altogether give 
$$ \int_{-\infty}^{\infty} \frac{dx}{(x^2 + 4 )^5} =(3+1/2)(2+1/2)(1+1/2)(1/2) \frac{\pi}{4! * 2.4^4} = \frac{35\pi}{4^8} = \frac{35\pi}{65536}$$
