get a integral from another

Question

if $\int\limits_{0}^{+\infty}x^3e^{-\alpha x^2} dx=\frac{1}{2A}$ then $\int\limits_{0}^{+\infty}x^4e^{-\alpha x^2} dx=$

i tried to use integration by parts $$\begin{align} \int\limits_{0}^{+\infty}x^4e^{-\alpha x^2} dx&=\int\limits_{0}^{+\infty}x^3 xe^{-\alpha x^2} dx\\ u=x^3&\Rightarrow du=3x^2dx\\ dv=xe^{-\alpha x^2}dx&\Rightarrow v=-\frac{1}{2\alpha}e^{-\alpha x^2}\\ \int\limits_{0}^{+\infty}x^4e^{-\alpha x^2} dx&=-\left.\frac{x^3}{2\alpha}e^{-\alpha x^2}\right|_{0}^{+\infty}+\frac{3}{2\alpha}\int\limits_{0}^{+\infty}x^2e^{-\alpha x^2}dx \end{align}$$ which appear not help much.

Answer 1

Outline: Use integration by parts to calculate $\int_0^\infty x^3 e^{-\alpha x^2}\,dx$. We know this is $\frac{1}{2A}$, so we can calculate $\alpha$.

Now use a couple of integrations by parts to calculate $\int_0^\infty x^4 e^{-\alpha x^2}\,dx$.

Not elegant, but it does the job.

Answer 2

Actually, you can get $$ \int_{0}^{+\infty} x^{2 n} e^{-a x^{2}} \mathrm{~d} x=\frac{\sqrt{\pi}}{2} \frac{(2 n-1) ! !}{2^{n}} a^{-\left(n+\frac{1}{2}\right)} $$ by integration by parts , where $a>0, n \in \mathbb{N}$