Find a recursive formula for $J_n = \int_0^{\infty} x^{2n+1}e^{-x^2} \, \mathrm{d}x$

Question

Given $$ J_n\stackrel{\text{def}}{=}\int_{0}^{+\infty}x^{2n+1}e^{-x^2}\,dx $$ compute $J_0$, $J_1$ and find a recursion formula for $J_n$.

I have computed $J_0$ and $J_1$ but I am clueless about the recursion formula, may I ask for some help? By integration by parts I got $$ J_0 = \frac{1}{2},\qquad J_1 = \frac{1}{2}.$$

Answer 1

By setting $x=\sqrt{z}$ we have

$$ J_n = \frac{1}{2}\int_{0}^{+\infty} z^{n} e^{-z}\,dz = \frac{n!}{2} $$ due to the integral definition of the $\Gamma$ function. In particular

$$ J_{n+1} = (n+1)\,J_n$$ holds by integration by parts.

Answer 2

Alternatively write $$\begin{align}J_n &= -\frac{1}{2}\int_0^{\infty}-2xe^{-x^2}x^{2n} \, \mathrm{d}x\\ & = -\frac{1}{2}\bigg[x^{2n}e^{-x^2}\bigg]_0^{\infty} + \frac{1}{2}\int_0^{\infty} 2nx^{2n-1}e^{-x^2} \, \mathrm{d}x \\ & = nJ_{n-1}\end{align}$$

using IBP with $u = x^{2n}$ and $\mathrm{d}v = -2xe^{-x^2} \implies v=e^{-x^2}$.