Integrate $x^2 e^{-x^2/2}$

Question

Is it possible to integrate $$\int_0^{\infty} x^2 e^{-x^2/2}\, \mathrm dx$$ by hand?

The answer is $\frac{1}{2\sqrt{2}}$ My apologies if this does not meet the standards of this blog. I will delete it if requested.

Answer 1

By the Feynman trick we have:

$$I = \lim_{a\to 1}\int_0^{+\infty} -2\left(\frac{\text{d}}{\text{d}a} e^{-(a x^2)/2}\right)\ \text{d}x = \lim_{a\to 1}-2\frac{\text{d}}{\text{d}a}\int_0^{+\infty} e^{-(ax^2)/2}\ \text{d}x = \lim_{a\to 1} -2 \frac{\text{d}}{\text{d}a}\sqrt{\frac{\pi}{2a}}$$

Hence

$$I = \lim_{a\to 1}-2\left(-\frac{1}{2} \sqrt{\frac{\pi }{2}} \left(\frac{1}{a}\right)^{3/2}\right)$$

And our integral is simply

$$I = \sqrt{\frac{\pi }{2}}$$

Which is the result of your integral.

Answer 2

Hint: $u = x$, $dv = xe^{-x^2/2}dx$

Answer 3

No tricks, just the Gamma integral: Substituting $x = \sqrt{2t}$ gives $$\int_0^\infty x^2 e^{-x^2/2} \,dx = \sqrt{2\vphantom{X}} \int_0^\infty t^{1/2} e^{-t} \,dt = \sqrt{2\vphantom{X}}\,\Gamma\Bigl(\frac32\Bigr) = \sqrt{\frac\pi2}.$$

Answer 4

Using a standard probability distribution:

If you know the Gaussian distribution $\mathcal{G}(\mu,\sigma)$: its pdf is $f_{\mu,\sigma}\colon\mathbb{R}\to\mathbb{R}$ defined by $$ f_{\mu,\sigma}(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(x-\mu)^2}{2\sigma^2}} $$ and you want to compute, for $X\sim\mathcal{G}(0,1)$, $$\begin{align*} \int_{0}^\infty x^2e^{-\frac{x^2}{2}}dx &= \frac{1}{2}\cdot\sqrt{2\pi}\int_{-\infty}^\infty x^2f_{0,1}(x)dx = \frac{\sqrt{2\pi}}{2} \mathbb{E}[X^2] = \frac{\sqrt{2\pi}}{2}\left( \mathbb{E}[X^2]-\mathbb{E}[X]^2\right) \\&= \frac{\sqrt{2\pi}}{2}\operatorname{Var}X = \frac{\sqrt{2\pi}}{2}\cdot 1 \\&= \sqrt{\frac{\pi}{2}} \end{align*}$$ where for the first step we used the fact that $x\mapsto x^2e^{-\frac{x^2}{2}}$ is an even function (hence the factor $\frac{1}{2}$ and the change of bounds in the integral).

Answer 5

$$\begin{align} u&=x^2/2 \\ \mathrm{d}u&=x\mathrm{d}x \\ \int x^2e^{-x^2/2}\,\mathrm{d}x &=\int \frac{x^2e^{-u}}{x}\,\mathrm{d}u \\&=\int \sqrt{2u}e^{-u}\,\mathrm{d}u \\&=\sqrt{2}\int u^{1/2}e^{-u}\,\mathrm{d}u \\\int_0^\infty x^2e^{-x^2/2}\,\mathrm{d}x&=\sqrt{2}\:\Gamma\left(\frac{1}{2}+1 \right)* \\&=\sqrt{2}\frac{\sqrt{\pi}}{2} \\&\quad \text{* where $\Gamma(z)$ is the gamma function $\int_0^\infty u^{z-1}e^u\mathrm{d}u$} \end{align}$$

Answer 6

As a first step note that $$ I=\int x e^{-\frac{x^2}{2}}dx $$ can be integrated with the substitution $$ -\frac{x^2}{2}=t \quad \rightarrow \quad xdx=-dt $$ and we have: $$ I=-\int e^t dt = -e^t+C=-e^{-\frac{x^2}{2}}+C $$

now you can write your integral as: $$ J=\int_0^{\infty} x^2 e^{-x^2/2}dx=\int_0^{\infty} x d\left(-e^{-\frac{x^2}{2}} \right) $$

and, integrating by part: $$ J=\left[-xe^{-x^2/2}\right]_0^{\infty}+\int_0^{\infty}e^{-x^2/2}dx $$

The first part is obviously $=0$. For the second, using the definition of the Error function, we have: $$ \int_0^{\infty}e^{-x^2/2}dx=\left[\sqrt{\frac{\pi}{2}}\mbox{erf}(x/\sqrt{2})\right] _0^{\infty} $$

and, since $\mbox{erf}(0)=0$ and $\lim _{x\to \infty}\mbox{erf}(x)=1$ we have $J=\sqrt{\frac{\pi}{2}}$.