# 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.

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

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.

• This is really a beautiful answer indeed. Thank you. 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. 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}

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


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}
• 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... Dec 5 '13 at 10:41
• @TonyK I just added the $\large =$ sign. Thanks. Dec 6 '13 at 2:27

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}

• You integrated $e^{-x^2}$ he wants $x^2 e^{-x^2}$. Dec 4 '13 at 18:54

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 $2!$.

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

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

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