TITLE: Integrate $x^2 e^{-x^2/2}$
QUESTION [6 upvotes]: Is it possible to integrate $$\int_0^{\infty} x^2 e^{-x^2/2}\, \mathrm dx$$
  by hand?

The answer is $\frac{1}{2\sqrt{2}}$ My apologies if this does not meet the standards of this blog. I will delete it if requested.

REPLY [4 votes]: Using a standard probability distribution:
If you know the Gaussian distribution $\mathcal{G}(\mu,\sigma)$: its pdf is $f_{\mu,\sigma}\colon\mathbb{R}\to\mathbb{R}$ defined by
$$
f_{\mu,\sigma}(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(x-\mu)^2}{2\sigma^2}}
$$
and you want to compute, for $X\sim\mathcal{G}(0,1)$,
$$\begin{align*}
\int_{0}^\infty x^2e^{-\frac{x^2}{2}}dx &= \frac{1}{2}\cdot\sqrt{2\pi}\int_{-\infty}^\infty x^2f_{0,1}(x)dx =  \frac{\sqrt{2\pi}}{2} \mathbb{E}[X^2] = \frac{\sqrt{2\pi}}{2}\left( \mathbb{E}[X^2]-\mathbb{E}[X]^2\right) \\&= \frac{\sqrt{2\pi}}{2}\operatorname{Var}X = \frac{\sqrt{2\pi}}{2}\cdot 1
\\&= \sqrt{\frac{\pi}{2}}
\end{align*}$$
where for the first step we used the fact that $x\mapsto x^2e^{-\frac{x^2}{2}}$ is an even function (hence the factor $\frac{1}{2}$ and the change of bounds in the integral).