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In Stewart's Calculus book I came across the following Gaussian integral.

Using $\int_{-\infty}^{\infty}\exp{(-x^2)}dx = \frac{\sqrt{\pi}}{2}$ evaluate $$ \int_0^{\infty} x^2 e^{-x^2}dx $$

I read in this pdf that

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

and how to use differentiation under the integral sign to evaluate it (recreated below for convenience).

$$ \begin{align*} I(a) &= \int_{-\infty}^{\infty} e^{-ax^2} dx =\sqrt{ \frac{\pi}{a}} \\ I'(a)&= -\int_{-\infty}^{\infty} x^2 e^{-ax^2}dx = -\frac{1}{2}\sqrt{\pi} a^{-3/2} \\I'(1) &= \frac{\sqrt{\pi}}{2} \end{align*} $$ Using the results above (and Wolfram Alpha), I was able to conclude that $$ \int_0^{\infty} x^2 e^{-x^2}dx = \frac{\sqrt{\pi}}{4} $$ however, I was wondering if there is some substitution or an another way to evaluate the aforementioned integral seeing as Leibniz's rule is not mentioned anywhere in the chapter.

rgarci0959
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3 Answers3

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Observe that $$\left(e^{-x^2}\right)'=-2xe^{-x^2} \implies xe^{-x^2}=-\frac12\left(e^{-x^2}\right)'$$ so that the integral can be written as $$-\frac12\int_{0}^{+\infty}x\left(e^{-x^2}\right)'dx$$ and integration by parts gives $$-\frac12\int_{0}^{+\infty}x\left(e^{-x^2}\right)'dx=-\frac12\left[xe^{-x^2}\right]_{0}^{+\infty}du+\frac12\int_{0}^{+\infty}e^{-x^2}dx$$ (notice the abuse of notation in the upper limit. It should be $\lim_{t\to \infty}$ instead of $\infty$). It remains to show that $\lim_{x\to \infty}xe^{-x^2}=0$ which can be done by L'Hopitals rule, since $$\lim_{x\to \infty}xe^{-x^2}=\lim_{x\to\infty}\frac{x}{e^{x^2}}\overset{\frac{\infty}{\infty}}=\lim_{x\to\infty}\frac{1}{2xe^{x^2}}=0$$

Jimmy R.
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Alternatively, you can use hueristics that stem from "probabilistic" results, i.e., note that if $X\sim \mathcal{N}(0,1/2)$, then its density function is given by $$ f_X(x) = \frac{1}{\sqrt{\pi}}e^{-x^2}, \quad x\in \mathbb{R} $$ hence, $\mathbb{V}(X)=\mathbb{E}(X^2) = \int_{-\infty}^{\infty}\frac{x^2}{\sqrt{\pi}}e^{-x^2}dx=1/2 $, thus by using symmetry of $f_X(x)$ w.r.t. $0$, \begin{align} \int_{-\infty}^{\infty}\frac{x^2}{\sqrt{\pi}}e^{-x^2}dx&= \int_{-\infty}^{0}\frac{x^2}{\sqrt{\pi}}e^{-x^2}dx + \int_{0}^{\infty}\frac{x^2}{\sqrt{\pi}}e^{-x^2}dx\\ &=\frac{2}{\sqrt{\pi}}\int_{0}^{\infty}x^2e^{-x^2}dx = 1/2 \end{align} rearranging the equation we get, $$ \int_{0}^{\infty}x^2e^{-x^2}dx = \frac{\sqrt{\pi}}{4}\, . $$

V. Vancak
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    It's very useful to know about that $\int x^ne^{-x^2}$ can be related to moments of the normal distribution. However it should be said that in practically every textbook the result that $V(X) = 1/2$ is proven by explicitly evaluating the integral by other means so as a derivation this is kind of circular. – Winther Aug 17 '16 at 11:56
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For $$\int\limits_{0}^{\infty} x^{2} \mathrm{e}^{-x^{2}} \mathrm{d} x$$ let $y = x^{2}$ \begin{equation} \int\limits_{0}^{\infty} x^{2} \mathrm{e}^{-x^{2}} \mathrm{d} x = \frac{1}{2} \int\limits_{0}^{\infty} \mathrm{e}^{-y} y^{\frac{1}{2}} \mathrm{d} y = \frac{1}{2} \Gamma\left(\frac{3}{2}\right) = \frac{\sqrt{\pi}}{4} \end{equation}

poweierstrass
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