Evaluate $\int_{-\infty}^{\infty}\frac{1}{(x^2+4)^5}dx$

Question

$$\int_{-\infty}^{\infty}\frac{1}{(x^2+4)^5}dx$$

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 $0$

Answer 1

Hint

Let $\Gamma_R=\{Re^{i\theta }\mid \theta \in [0,\pi]\}\cup[-R,R]$. Then, $$\int_{\Gamma_R} \frac{1}{(z^2+4)^5}=Res_{z=2i}(f)=\lim_{z\to 2i}\frac{1}{4!}\frac{d^4}{dz^4}\frac{1}{(z+2i)^5}=...$$

Answer 2

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

$$ \int_{-\infty}^{\infty} \frac{dx}{(x^2 + 4 )^5} =(3+1/2)(2+1/2)(1+1/2)(1/2) \frac{\pi}{4! * 2.4^4} = \frac{35\pi}{4^8} = \frac{35\pi}{65536}$$

Answer 3

$f(x)=\dfrac{1}{(x^2+4)^5}$

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 $2\pi i\, \mathrm {Res} (f,2i)=\dfrac{35 \pi }{65536}$

Answer 4

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: $$ \int_{-\infty}^{+\infty}\frac{dx}{(x^2+4)^5} = \color{red}{\frac{35 \pi}{2^{16}}}.$$