Integrate $x^2 e^{-x^2/2}$ 
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.
 A: Hint: $u = x$, $dv = xe^{-x^2/2}dx$
A: No tricks, just the Gamma integral: Substituting $x = \sqrt{2t}$ gives
$$\int_0^\infty x^2 e^{-x^2/2} \,dx = \sqrt{2\vphantom{X}} \int_0^\infty t^{1/2} e^{-t} \,dt = \sqrt{2\vphantom{X}}\,\Gamma\Bigl(\frac32\Bigr) = \sqrt{\frac\pi2}.$$
A: 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).
A: $$\begin{align}
u&=x^2/2
\\
\mathrm{d}u&=x\mathrm{d}x
\\
\int x^2e^{-x^2/2}\,\mathrm{d}x
&=\int \frac{x^2e^{-u}}{x}\,\mathrm{d}u
\\&=\int \sqrt{2u}e^{-u}\,\mathrm{d}u
\\&=\sqrt{2}\int u^{1/2}e^{-u}\,\mathrm{d}u
\\\int_0^\infty x^2e^{-x^2/2}\,\mathrm{d}x&=\sqrt{2}\:\Gamma\left(\frac{1}{2}+1 \right)*
\\&=\sqrt{2}\frac{\sqrt{\pi}}{2}
\\&\quad \text{* where $\Gamma(z)$ is the gamma function $\int_0^\infty u^{z-1}e^u\mathrm{d}u$}
\end{align}$$
A: By the Feynman trick we have:
$$I = \lim_{a\to 1}\int_0^{+\infty} -2\left(\frac{\text{d}}{\text{d}a} e^{-(a x^2)/2}\right)\ \text{d}x = \lim_{a\to 1}-2\frac{\text{d}}{\text{d}a}\int_0^{+\infty} e^{-(ax^2)/2}\ \text{d}x = \lim_{a\to 1} -2 \frac{\text{d}}{\text{d}a}\sqrt{\frac{\pi}{2a}}$$
Hence
$$I = \lim_{a\to 1}-2\left(-\frac{1}{2} \sqrt{\frac{\pi }{2}} \left(\frac{1}{a}\right)^{3/2}\right)$$
And our integral is simply
$$I = \sqrt{\frac{\pi }{2}}$$
Which is the result of your integral.
A: As a first step note that 
$$
I=\int x e^{-\frac{x^2}{2}}dx
$$
can be integrated with the substitution
$$
-\frac{x^2}{2}=t \quad \rightarrow \quad xdx=-dt
$$
and we have:
$$
I=-\int e^t dt = -e^t+C=-e^{-\frac{x^2}{2}}+C
$$
now you can write your integral as:
$$
J=\int_0^{\infty} x^2 e^{-x^2/2}dx=\int_0^{\infty} x d\left(-e^{-\frac{x^2}{2}} \right)
$$
and, integrating by part:
$$
J=\left[-xe^{-x^2/2}\right]_0^{\infty}+\int_0^{\infty}e^{-x^2/2}dx
$$
The first part is obviously $=0$. For the second, using the definition of the Error function, we have:
$$
\int_0^{\infty}e^{-x^2/2}dx=\left[\sqrt{\frac{\pi}{2}}\mbox{erf}(x/\sqrt{2})\right] _0^{\infty}
$$ 
and, since $\mbox{erf}(0)=0$ and $\lim _{x\to \infty}\mbox{erf}(x)=1$ we have $J=\sqrt{\frac{\pi}{2}}$.
