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 $0$
 A: Hint
Let $\Gamma_R=\{Re^{i\theta }\mid \theta \in [0,\pi]\}\cup[-R,R]$. Then, $$\int_{\Gamma_R} \frac{1}{(z^2+4)^5}=Res_{z=2i}(f)=\lim_{z\to 2i}\frac{1}{4!}\frac{d^4}{dz^4}\frac{1}{(z+2i)^5}=...$$
A: $f(x)=\dfrac{1}{(x^2+4)^5}$
The pole $x=2i$ is $5-$ple so you must use this formula
$$\mathrm {Res} (f,2i)={\frac {1}{(5-1)!}}\lim _{z\to 2i}{\frac {d^{4}}{dx^{4}}}\left(\dfrac{(x-2 i)^5}{(x+2 i)^5 (x-2 i)^5}\right)$$
$$\mathrm {Res} (f,2i)={\frac {1}{24}}\lim _{z\to 2i}\frac{1680}{(x+2 i)^9}=-\frac{35 i}{131072}$$
So the integral is $2\pi i\, \mathrm {Res} (f,2i)=\dfrac{35 \pi }{65536}$
A: For any $a>0$ we have
$$ \int_{-\infty}^{+\infty}\frac{dx}{x^2+a}=\frac{\pi}{\sqrt{a}} $$
and by applying $\frac{d^4}{da^4}$ to both sides we get:
$$ 24\int_{-\infty}^{+\infty}\frac{dx}{(x^2+a)^5}=\frac{105 \pi}{16 a^4\sqrt{a}} $$
so by rearranging and evaluating at $a=4$ we get:
$$ \int_{-\infty}^{+\infty}\frac{dx}{(x^2+4)^5} = \color{red}{\frac{35 \pi}{2^{16}}}.$$
A: 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 
$$ \int_{-\infty}^{\infty} \frac{dx}{(x^2 + 4 )^5} =(3+1/2)(2+1/2)(1+1/2)(1/2) \frac{\pi}{4! * 2.4^4} = \frac{35\pi}{4^8} = \frac{35\pi}{65536}$$
