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 ?
$\newcommand{\H}{\mathscr{H}}$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.