Improper integral

Let $n \in \mathbb{N}$. Evaluate the integral:

$$\int_0^\infty \frac{{\rm d}x}{(1+x^2)^n}$$

Solution:



Successively we have:

$$\begin{aligned}
\int_{0}^{\infty}\frac{{\rm d}x}{\left ( 1+x^2 \right )^n} &\overset{u=x^2}{=\! =\! =\!}\frac{1}{2}\int_{0}^{\infty} \frac{{\rm d}u}{\sqrt{u}\left ( 1+u \right )^n}\\
 &= \frac{1}{2}\int_{0}^{\infty}\frac{u^{1/2-1}}{\left ( 1+u \right )^{1/2 +(n-1/2)}}\,{\rm d}u\\
 &= \frac{1}{2}{\rm B}\left ( \frac{1}{2}, n-\frac{1}{2} \right )\\
 &= \frac{1}{2}\cdot \frac{\Gamma \left ( \frac{1}{2} \right )\Gamma \left ( n-\frac{1}{2} \right )}{\Gamma (n)} \\
 &=\frac{1}{2}\cdot \frac{\sqrt{\pi}\Gamma \left ( n-\frac{1}{2} \right )}{(n-1)!}
\end{aligned}$$

Recalling $\Gamma$'s functional equation and subbing $x$ with $n-\dfrac{1}{2}$ we get that:

 $$\Gamma \left ( n-\frac{1}{2} \right )=\frac{1}{n-\frac{1}{2}}\Gamma \left ( n+\frac{1}{2} \right )=\frac{(2n)!\sqrt{\pi}}{\left( n- \dfrac{1}{2} \right) 4^n n!}$$

Hence:

$$ \int_{0}^{\infty}\frac{{\rm d}x}{\left ( x^2+1 \right )^n }=\frac{\left ( 2n \right )! \pi}{2\cdot 4^n n! (n-1)! \left ( n-\frac{1}{2} \right )}, \;\; n \in \mathbb{N}$$

and the exercise comes to an end.