Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I recognize that the $\int_0^\infty \mathrm{e}^{-x}x^n\,\mathrm{d}x = \Gamma(n+1)$ and $\int_{-\infty}^{+\infty} \mathrm{e}^{-x^2}\,\mathrm{d}x = \sqrt{\pi}$. I am having difficulty, however with $\int_{-\infty}^{+\infty} \mathrm{e}^{-x^2}x^n\,\mathrm{d}x$. By the substitution $u=x^2$, this can be equivalently expressed as $\frac{1}{2} \int_{-\infty}^{+\infty} \mathrm{e}^{-u}u^{\frac{n-1}{2}}\,\mathrm{d}u$. This integral is similar to the first one listed (which equates to the $\Gamma$ function), except that its domain spans $\mathbb{R}$ like the second integral (which equates to $\sqrt{\pi}$). Any pointers on how to evaluate this integral would be helpful.

share|improve this question
    
See here to see how to do the even case. For the odd case integrate by parts with $u= x$ and $dv = \mbox{the rest}$ and result will follow from knowing the even case. –  Ragib Zaman Feb 19 '12 at 12:55
    
Ahh, didn't realize the integration was over the entire real line. As Davide shows, it's simpler than the wiki link. –  Ragib Zaman Feb 19 '12 at 13:30

3 Answers 3

up vote 3 down vote accepted

Let $I_n:=\int_{-\infty}^{+\infty}e^{-x^2}x^ndx$. If $n$ is odd then $I_n=0$ and for $p\geq 1$: \begin{align} I_{2p}&=\int_0^{+\infty}e^{-x^2}x^{2p}dx+\int_{-\infty}^0e^{-x^2}x^{2p}dx\\ &=\int_0^{+\infty}e^{-t^2}t^{2p}dt+\int_0^{+\infty}e^{-t^2}(-t)^{2p}dt\quad (\mbox{left: } t=x,\mbox{right: } t=-x)\\ &=2\int_0^{+\infty}e^{-t^2}t^{2p}dt\\ &=2\int_0^{+\infty}e^{-s}s^p\frac 1{2\sqrt s}ds \quad (s=t^2)\\ &=\int_0^{+\infty}e^{-s}s^{p-1/2}ds\\ &=\left[-e^{-s}s^{p-1/2}\right]_0^{+\infty}+\int_0^{+\infty}e^{—s}\left(p-\frac 12\right)s^{p-1-1/2ds}\\ &=\left(p-\frac 12\right)I_{2(p-1)}. \end{align} Finally we get $I_{2p+1}=0$ and $I_{2p}=\sqrt \pi\prod_{j=1}^p\left(j-\frac 12\right)$ for all $p\geq 0$.

share|improve this answer
    
Thanks, Davide. Should the last differential in the second line be $\mathrm{d}t$ instead of $\mathrm{d}x$? To go from the first line to the second line, did you simply apply the substitutions $x=t$ (left) and $x=-t$ (right)? –  user001 Feb 19 '12 at 13:26
    
The solution is a bit different from the one in the wikipedia link posted by @RagibZaman, which takes the form $I_2n = \int_{-\infty}^{+\infty} e^{-x^2} x^{2n} \,\mathrm{d}x = \frac{\sqrt{\pi}}{2^n}(2n-1)!!$. I am probably missing something. –  user001 Feb 19 '12 at 13:33
    
In fact I make a mistake in the last substitution. –  Davide Giraudo Feb 19 '12 at 13:39
    
Thank you, Davide. Where you have written $\int_0^{+\infty}e^{-t^2}(-t)^{2p}dx$ in the second line, should the $\mathrm{d}x$ be $\mathrm{d}t$ instead? Also, is the substitution for the first term on that line $t=x$ (i.e., you used different substitutions for the left and right terms)? Thank you for clarifying. –  user001 Feb 19 '12 at 13:48
    
@user001 Thanks, it's done now. –  Davide Giraudo Feb 19 '12 at 13:58

The function $x\mapsto x^n e^{-x^2}$ is absolutely integrable on the real line. If $n$ is odd, the integrand is odd, and we have $$\int_0^\infty x^n e^{-x^2}\, dx = 0$$

Now consider the even case. We first use symmetry to get the integral onto $[0,\infty)$ and then use the subsitution $x \rightarrow \sqrt{x}$ as follows $$\int_{-\infty}^\infty x^n e^{-x^2}\, dx= 2\int_0^\infty x^n e^{-x^2}\, dx = 2\int_0^\infty x^{n/2} e^{-x}{dx\over2\sqrt{x}} = \Gamma\left({n + 1\over 2}\right).$$

Invoking the factorial property of the $\Gamma$ function relates this solution to the other posted solution.

share|improve this answer
    
Thank you very much for mentioning that the solution distills down to the $\Gamma$ function. I think the argument of this function should be $(n+1)/2$ rather than $n+1/2$. –  user001 Feb 19 '12 at 14:08
    
Thanks, '001. I fixed that. –  ncmathsadist Feb 19 '12 at 18:54

A very simple way to find the integral is to define

$$I(a) \equiv \int_{-\infty}^\infty e^{-ax^2}dx = \sqrt{\frac{\pi}{a}}$$

then we have

$$\frac{d^kI(a)}{da^k} = (-1)^k\int_{-\infty}^\infty x^{2k}e^{-ax^2}dx = \sqrt{\pi}\frac{d^k}{da^k}a^{-1/2}$$

Taking the $k$'th derivative of $a^{-1/2}$ and putting $a=1$ gives us

$$\int_{-\infty}^\infty x^{2k}e^{-x^2}dx = \sqrt{\pi}\frac{(2k)!}{4^k k!}$$

This method is sometimes called Feynman integration.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.