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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.

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6  
Try integration by parts. –  universalset Dec 4 '13 at 18:10
2  
$x^2e^{-x^2} = \frac x2\cdot 2xe^{-x^2}$ –  Greg Martin Dec 4 '13 at 18:14
    
use parts,it is helpfull –  Half-Blood prince Dec 5 '13 at 11:06

7 Answers 7

up vote 22 down vote accepted

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.

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This is really a beautiful answer indeed. Thank you. –  Vaaal Dec 6 '13 at 13:22
    
I still remember the day when one of my classmates showed me this and told me he learned this particular trick from one of my favorite professors. –  nayrb Dec 6 '13 at 14:08

$$\begin{align} \int_{-\infty}^\infty x^2e^{-x^2}\text dx&=2\int_0^\infty x^2e^{-x^2}\text dx\\ &=\int_0^\infty \sqrt ue^{-u}\text du\\ &=\Gamma\left(\frac32\right)\\ &=\frac12\Gamma\left(\frac12\right)\\ &=\frac12\int_0^\infty\frac{e^{-u}}{\sqrt u}\text du\\ &=\int_0^\infty e^{-x^2}\text dx\\ &=\frac{\sqrt \pi}2 \end{align}$$

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Can you show how to get there without using the gamma function? –  Vaaal Dec 4 '13 at 19:24
    
Is that more acceptable? –  Tim Ratigan Dec 4 '13 at 20:30

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: $$\begin{align} \int_{-\infty}^{\infty}x^{2}e^{-x^{2}}dx&=\int_{-\infty}^{\infty}x\frac{d}{dx}\left[-\frac{1}{2}e^{-x^{2}}\right]dx = \left.x\left[-\frac{1}{2}e^{-x^{2}}\right]\right|_{-\infty}^{\infty}+\frac{1}{2}\int_{-\infty}^{\infty}e^{-x^{2}}dx\\ &= \frac{1}{2}\int_{-\infty}^{\infty}e^{-x^{2}}dx=\frac{\sqrt{\pi}}{2}.\end{align} $$

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1  
You integrated $e^{-x^2}$ he wants $x^2 e^{-x^2}$. –  Deven Ware Dec 4 '13 at 18:54
3  
Read his full answer! –  nayrb Dec 5 '13 at 22:12

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

  1. \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} $$\color{#0000ff}{\large% \int_{-\infty}^{\infty}x^{2}\expo{-x^{2}}\,\dd x = {\root{\pi} \over 2}} $$
  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}
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I think the first line is missing an '=' sign, is that right? I was going to edit it myself until I saw your $L_AT_EX$ hell... –  TonyK Dec 5 '13 at 10:41
    
@TonyK I just added the $\large =$ sign. Thanks. –  Felix Marin Dec 6 '13 at 2:27

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 $2!$.

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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 $1/\sqrt{1-4t} = \sum_{n=0}^\infty {2n \choose n}t^n$).

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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

$$\int_{-\infty}^\infty x^2 e^{-x^2}\text dx= -\frac{x}{2}e^{-x^2}\bigg|_{-\infty}^\infty+\frac{1}{2}\int_{-\infty}^\infty e^{-x^2}dx=0-0+\frac{1}{2}\sqrt{\pi}=\frac{\sqrt{\pi}}{2}$$

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