About an equivalent of $\displaystyle \int_{0}^{+\infty}\frac{\text{d}x}{\left(\alpha^2+x^2\right)^n}$ [ Checking a result ] I like creating exercise or just practicing some math stuff and I was to study for $n \in \mathbb{N}^{*}$ the sequence
$$
\mathscr{H}_n=\int_{0}^{+\infty}\frac{\text{d}x}{\left(\alpha^2+x^2\right)^n}
$$
where $\alpha \in \mathbb{R}^{*+}$.
I've shown that for all $n \in \mathbb{N}^{*}$
$$
\mathscr{H}_n=\frac{\pi}{\left(2\alpha\right)^{2n-1}}\frac{\left(2n-2\right)!}{\left(n-1\right)!^2}
$$
Using Stirling Formula, I think I've shown that
$$
\frac{\left(2n-2\right)!}{\left(n-1\right)!^2}\underset{(+\infty)}{\sim}\frac{4^{n-1}}{\sqrt{\pi\left(n-1\right)}}
$$
But I'm not sure of this and i've tested for $n=70$ and it seems like okay but I would like to know if i'm right. Because it would imply that

$$
\mathscr{H}_n\underset{(+\infty)}{\sim}\frac{1}{2\alpha^{2n-1}}\sqrt{\frac{\pi}{n}}
$$

which would be nice. Can somebody tells me if the results I found are correct ?
 A: $\newcommand{\H}{\mathscr{H}}$For the asymptotics you don't need to evaluate the intgeral. One has:
$$\H_n=\int^\infty_0 e^{-n\ln(x^2+a^2)}\,dx$$
As $n\to \infty$ the main contribution is near $0$ since $\ln(x^2+a^2)$ is strictly increasing. I'll use Taylor approximation around $x=0$ of second order to find the asymptotics (the first order does not give any "information"). That is:
\begin{align}
\ln(x^2+a^2)=\ln(a^2)+\frac{x^2}{a^2}+O(x^3) \ \ \  \ \text{ as }\  x\to 0
\end{align} This all can be made rigorous using aysmptotics analysis which I will not do here, but I'll just write it down:
\begin{align}
\H_n \sim \int^\infty_0 e^{-n(\ln(a^2)+x^2/a^2)}\,dx=e^{-n\ln(a^2)}\int^\infty_0 e^{-nx ^2/a^2}\,dx=\frac{1}{a^{2n}}\frac{a\sqrt[]{\pi}}{2\sqrt[]{n}}=\frac{1}{2a^{2n-1}}\sqrt[]{\frac{\pi}{n}}
\end{align}
The proof for the Laplace method can help you to prove this rigorously. 
A: Based on looking up an integral  I get your integral $=\frac{1}{\alpha^{2n-1}}\int_0^{\infty} \frac{dx}{(1+x^2)^n}=\frac{1}{2\alpha^{2n-1}}\frac{\Gamma (\frac{1}{2})\Gamma (n-\frac{1}{2})}{\Gamma (n)}$
