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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.

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  • $\begingroup$ 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. $\endgroup$ Feb 19, 2012 at 12:55
  • $\begingroup$ Ahh, didn't realize the integration was over the entire real line. As Davide shows, it's simpler than the wiki link. $\endgroup$ Feb 19, 2012 at 13:30

2 Answers 2

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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.

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  • $\begingroup$ 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 $n+1/2$. $\endgroup$
    – user001
    Feb 19, 2012 at 14:08
  • $\begingroup$ Thanks, '001. I fixed that. $\endgroup$ Feb 19, 2012 at 18:54
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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 $p\geq 0$.

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  • $\begingroup$ 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)? $\endgroup$
    – user001
    Feb 19, 2012 at 13:26
  • $\begingroup$ 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. $\endgroup$
    – user001
    Feb 19, 2012 at 13:33
  • $\begingroup$ In fact I make a mistake in the last substitution. $\endgroup$ Feb 19, 2012 at 13:39
  • $\begingroup$ 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. $\endgroup$
    – user001
    Feb 19, 2012 at 13:48
  • $\begingroup$ @user001 Thanks, it's done now. $\endgroup$ Feb 19, 2012 at 13:58

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