integral of $x^2e^{-x^2}~dx$ from $-\infty$ to $+\infty$ I know that the 
$$\int^{+\infty}_{-\infty}e^{-x^2}~dx$$ is equal to $\sqrt\pi$
It's also very clear that 
$$\int^{+\infty}_{-\infty}xe^{-x^2}~dx$$ is equal to 0;
However, I cannot manage to calculate this really similar integral.
$$\int^{+\infty}_{-\infty}x^2e^{-x^2}~dx$$
I know that the result is $\frac{\sqrt\pi}{2}$ but I don't know how to get to this result. I tried different substitution, but it doesn't seem to help. 
Any idea?
Thank you very much.
 A: $$\begin{align} \int_{-\infty}^\infty x^2e^{-x^2}\text dx&=2\int_0^\infty x^2e^{-x^2}\text dx\\
&=\int_0^\infty \sqrt ue^{-u}\text du\\
&=\Gamma\left(\frac32\right)\\
&=\frac12\Gamma\left(\frac12\right)\\
&=\frac12\int_0^\infty\frac{e^{-u}}{\sqrt u}\text du\\
&=\int_0^\infty e^{-x^2}\text dx\\
&=\frac{\sqrt \pi}2 \end{align}$$
A: $\newcommand{\+}{^{\dagger}}%
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*
*
\begin{align}
\int_{-\infty}^{\infty}x^{2}\expo{-x^{2}}\,\dd x
&=\left.-\,\partiald{}{\mu}\int_{-\infty}^{\infty}\expo{-\mu x^{2}}\,\dd x
\right\vert_{\,\mu\ =\ 1}
=
\left.-\,\partiald{}{\mu}\pars{\mu^{-1/2}\int_{-\infty}^{\infty}\expo{-x^{2}}\,\dd x}
\right\vert_{\,\mu\ =\ 1}
\\[3mm]&=
\left.-\pars{-\,\half}\mu^{-3/2}\right\vert_{\,\mu\ =\ 1}\
\underbrace{\quad\int_{-\infty}^{\infty}\expo{-x^{2}}\,\dd x\quad}
_{\ds{=\ \root{\pi}}}
\end{align}
$$\color{#0000ff}{\large%
\int_{-\infty}^{\infty}x^{2}\expo{-x^{2}}\,\dd x = {\root{\pi} \over 2}}
$$

*
\begin{align}
\color{#0000ff}{\large\int_{-\infty}^{\infty}x^{2}\expo{-x^{2}}\,\dd x}
&=\int_{-\infty}^{\infty}\pars{-\,\half}x\,\totald{\expo{-x^{2}}}{x}\,\dd x
=\overbrace{\left.-\,\half\,x\expo{-x^{2}}\right\vert
_{-\infty}^{\infty}}^{\ds{=\ 0}}
+
{1 \over 2}\int_{-\infty}^{\infty}\expo{-x^{2}}\,\dd x
\\[3mm]&= \color{#0000ff}{\large{\root{\pi} \over 2}}
\end{align}

A: The standard solution requires a trick using polar coordinates. If the value of your $\int_{-\infty}^{\infty}$ is $V$, then
$$
    V^{2} = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-x^{2}}e^{-y^{2}}dx dy = \int_{0}^{2\pi}\int_{0}^{\infty}e^{-r^{2}}rdrd\theta=2\pi\int_{0}^{\infty}e^{r^{2}}rdr=\pi.
$$
So $V = \int_{-\infty}^{\infty}e^{-x^{2}}dx=\sqrt{\pi}$. Integrating by parts finishes the problem as:
$$\begin{align}
   \int_{-\infty}^{\infty}x^{2}e^{-x^{2}}dx&=\int_{-\infty}^{\infty}x\frac{d}{dx}\left[-\frac{1}{2}e^{-x^{2}}\right]dx = \left.x\left[-\frac{1}{2}e^{-x^{2}}\right]\right|_{-\infty}^{\infty}+\frac{1}{2}\int_{-\infty}^{\infty}e^{-x^{2}}dx\\
    &= \frac{1}{2}\int_{-\infty}^{\infty}e^{-x^{2}}dx=\frac{\sqrt{\pi}}{2}.\end{align}
$$
A: Same trick as T.A.E. but without integrating by parts:
$$
\begin{align}
\left(\int_{-\infty}^\infty x^2\,e^{-x^2}\,\mathrm{d}x\right)^2
&=\int_{-\infty}^\infty\int_{-\infty}^\infty x^2y^2\,e^{-x^2}e^{-y^2}\,\mathrm{d}x\,\mathrm{d}y\\
&=\int_0^\infty\int_0^{2\pi}r^4\cos^2(\theta)\sin^2(\theta)\,e^{-r^2}\,r\,\mathrm{d}\theta\,\mathrm{d}r\\
&=\int_0^\infty r^4e^{-r^2}\,r\,\mathrm{d}r\int_0^{2\pi}\cos^2(\theta)\sin^2(\theta)\,\mathrm{d}\theta\\
&=\frac18\int_0^\infty s^2e^{-s}\,\mathrm{d}s\int_0^{2\pi}\sin^2(2\theta)\,\mathrm{d}\theta\\
&=\frac18\cdot2!\cdot\pi\\
&=\frac\pi4
\end{align}
$$
Therefore,
$$
\int_{-\infty}^\infty x^2\,e^{-x^2}\,\mathrm{d}x=\frac{\sqrt\pi}{2}
$$
One might use integration by parts to get $\int_0^\infty s^2e^{-s}\,\mathrm{d}s=2$ if one didn't recognize it as $2!$.
A: For $n\geq 0$ let $A_n = \int_{-\infty}^{\infty} x^{2n}e^{-x^2} dx$. Then
$$A_n =  \sqrt{\pi}\frac{{2n \choose n}n!}{4^n}.$$
Indeed, $$\sum_{n=0}^\infty A_n t^n/n! = \int_{-\infty}^\infty \sum_{n=0}^\infty e^{-x^2}(tx^2)^n dx/n! = \int_{-\infty}^\infty e^{-x^2(1-t)}dx = \frac{\sqrt \pi}{\sqrt{1-t}}.$$
Comparing coefficients of $t$ on both sides we find the above formula (using the fact that $1/\sqrt{1-4t} = \sum_{n=0}^\infty {2n \choose n}t^n$).
A: I'm surprised no one has given this answer.
We have of course by $u$-substitution
$$\int_{-\infty}^{\infty} e^{-\alpha x^2} dx = \sqrt{\frac{\pi}{\alpha}}.$$
Take a derivative of each side with respect to $\alpha$ to get
$$-\int_{-\infty}^\infty x^2 e^{-\alpha x^2} dx = -\frac{1}{2} \frac{\sqrt{\pi}}{\alpha^{3/2}}.$$
Substitute $\alpha = 1$ and cancel negative signs.
EDIT: I see that Felix Marin does essentially the same thing, but I think this is a better explanation.
A: Given that you know $\int_{-\infty}^\infty e^{-x^2}dx=\sqrt{\pi}$:
First, we note it's easy to integrate $x e^{-x^2}$ ( by substitution $u=x^2$)   $$\int x e^{-x^2}dx= -\frac{1}{2}e^{-x^2}$$
Then, we can apply integration by parts:
$$\int x \,x e^{-x^2}\text dx= -\frac{x}{2}e^{-x^2} + \frac{1}{2}\int e^{-x^2}dx$$
Therefore
$$\int_{-\infty}^\infty x^2  e^{-x^2}\text dx=  -\frac{x}{2}e^{-x^2}\bigg|_{-\infty}^\infty+\frac{1}{2}\int_{-\infty}^\infty e^{-x^2}dx=0-0+\frac{1}{2}\sqrt{\pi}=\frac{\sqrt{\pi}}{2}$$ 
