6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Try to do it by parts twice...or even once. $\endgroup$ – DonAntonio Sep 30 '16 at 18:46
  • $\begingroup$ I think you might be able to compare it to the gamma function? $\endgroup$ – Jam Sep 30 '16 at 18:47
  • $\begingroup$ ok, I am little rusty with calculus what do we set our u and dv at? $\endgroup$ – Wolfy Sep 30 '16 at 18:47
  • $\begingroup$ It seems that your answer is wrong: see wolframalpha.com/input/?i=integrate+x^2+e^%28-x^2%2F2%29+from+x%3D0+to+infty $\endgroup$ – Emilio Novati Sep 30 '16 at 19:19
14
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Very nice trick! $\endgroup$ – Babak Sep 30 '16 at 19:06
  • 1
    $\begingroup$ @Babak Thank you! I really love that trick too, and finally I could use it :D $\endgroup$ – Von Neumann Sep 30 '16 at 19:16
  • $\begingroup$ Nitpick: the LHS should not be $I(a)$ but $I$: it does not depend on $a$. Or you want to drop the $\lim_{a\to 1}$ there? $\endgroup$ – Clement C. Sep 30 '16 at 19:31
  • 1
    $\begingroup$ @FourierTransform +1 Feynman's Trick comes to the rescue for all of the even powered moments. $\endgroup$ – Mark Viola Sep 30 '16 at 20:00
  • 1
    $\begingroup$ @Lovsovs I forgot it, thanks! Going to edit $\endgroup$ – Von Neumann Sep 30 '16 at 20:26
7
$\begingroup$

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

$\endgroup$
5
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ Actually the Gamma function is a trick too :) But I give +1 because I love the Gamma function! $\endgroup$ – Von Neumann Sep 30 '16 at 19:17
  • $\begingroup$ And how does one evaluate the Gamma function at half integers? Using the functional relationship $\Gamma(x+1)=x\Gamma(x)$ boils it down to evaluating $\Gamma(1/2)$. And would you do that? $\endgroup$ – Mark Viola Sep 30 '16 at 19:55
  • $\begingroup$ @Dr.MV For this we can evaluate the Gaussian integral $\int_0^\infty e^{-x^2}\,dx = \sqrt\pi/2$ with another method, and then substituting $x = \sqrt{t}$ gives $\int_0^\infty e^{-x^2}\,dx = \Gamma(1/2)/2$. $\endgroup$ – arkeet Sep 30 '16 at 20:00
  • $\begingroup$ @arkeet Yes, I know that. I just want others to be aware that the evaluation of the integral of interest using the Gamma function requires that one evaluate the Gamma function. $\endgroup$ – Mark Viola Sep 30 '16 at 20:02
  • $\begingroup$ Yes, that is the point. But if you want to avoid the function, the integral can be reduced to the Gaussian integral more directly by grixor's answer. $\endgroup$ – arkeet Sep 30 '16 at 20:04
3
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ Clement, how are you my friend. This is a fine approach. But it might prompt, of course, one to ask, "how does one calculate the variance (or any moment) of the normal (Gaussian) distribution in terms of the parameters $\mu$ and $\sigma$ of the corresponding density function?" $\endgroup$ – Mark Viola Sep 30 '16 at 19:59
  • $\begingroup$ Indeed, it is circular (and I fully admit it). But assuming the variance and expectation of a Gaussian are known (which only requires to compute the integral once in a lifetime), this allows one to (re)derive such results very cheaply. $\endgroup$ – Clement C. Sep 30 '16 at 20:03
  • $\begingroup$ And that is the rationale for my giving a +1, well deservedly! $\endgroup$ – Mark Viola Sep 30 '16 at 20:04
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ And how does one evaluate the Gamma function at half integers? Using the functional relationship $\Gamma(x+1)=x\Gamma(x)$ boils it down to evaluating $\Gamma(1/2)$. And would you do that? $\endgroup$ – Mark Viola Sep 30 '16 at 19:55
0
$\begingroup$

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 $J=\sqrt{\frac{\pi}{2}}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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