I'm trying to evaluate the integral

$$ \int_0^\infty r^2 e^{-r^2/2\sigma^2} dr$$

Wolfram Alpha gives the answer

$$ \frac{\sqrt{\frac{\pi}{2}}}{(\frac{1}{\sigma^2})^{3/2}} $$

However, when I evaluated the integral it's undefined.

Let $u = r^2$ and $dv = e^{-r^2/2\sigma^2} dr$. Then $du = 2r \ dr$ and $v = - \frac{\sigma^2}{r} e^{-r^2/2\sigma^2}$.

$$ \int_0^\infty r^2 e^{-r^2/2\sigma^2} dr = \int_0^\infty u \ dv = u v\Big|^\infty_0 - \int_0^\infty v \ du $$

$$= \Big{[}- \sigma^2 r e^{-r^2/2\sigma^2} \Big{]}^\infty_0 + \int_0^\infty 2 \sigma^2 e^{-r^2/2\sigma^2} dr $$

For the first term,

$$ \lim_{r \to \infty} - \sigma^2 r e^{-r^2/2\sigma^2} = \lim_{r \to \infty} \frac{- \sigma^2 r }{e^{r^2/2\sigma^2}}$$

Since the numerator and denominator both tend to infinity, by l'Hospital's Rule,

$$ \stackrel{H}{=} \lim_{r \to \infty} \frac{- \sigma^2 }{\frac{r}{\sigma^2}e^{r^2/2\sigma^2}} = 0$$

For the other limit of the first term,

$$ \lim_{r \to 0} - \sigma^2 r e^{-r^2/2\sigma^2} = - \sigma^2 (0) (1) = 0 $$

So the first term is zero which means

$$ \int_0^\infty r^2 e^{-r^2/2\sigma^2} dr = \int_0^\infty 2 \sigma^2 e^{-r^2/2\sigma^2} dr = 2 \sigma^2 \Big{[}\frac{-\sigma^2}{r}e^{-r^2/2\sigma^2}\Big{]}^\infty_0$$

$$ = 2 \sigma^2 \Big{[}\lim_{r \to \infty}\frac{-\sigma^2}{r e^{r^2/2\sigma^2}} + \lim_{r \to 0}\frac{\sigma^2}{r e^{r^2/2\sigma^2}} \Big{]} = 2 \sigma^2 [0 \pm \infty]. $$

Any help is appreciated.

  • 5
    $\begingroup$ The integral of $\exp(-r^2/2\sigma^2)$ is not $-\sigma^2/r \cdot \exp(-r^2/2\sigma^2)$ $\endgroup$ – πr8 Dec 21 '16 at 19:02

Your error is to consider that $$ dv = e^{-r^2/2\sigma^2} dr \implies v = - \frac{\sigma^2}{r} e^{-r^2/2\sigma^2} $$ which is wrong as one may see by differentiating the latter expression.

If you want to make an integration by parts, you can rather write $$ u=r,\quad dv = re^{-r^2/2\sigma^2} dr, \quad du=dr,\quad v = -\frac1{\sigma^2}e^{-r^2/2\sigma^2}. $$

  • $\begingroup$ Ahh that makes sense. Thanks! $\endgroup$ – Y. Sargis Dec 21 '16 at 19:41
  • $\begingroup$ @Y.Sargis You are welcome. $\endgroup$ – Olivier Oloa Dec 21 '16 at 19:42


The way to fix it is to pick $dv = re^{-r^2/2\sigma^2}dr$ and let $u = r$


You dont choose the correct term in the integral by parts . You should pick $u=r$ and $v'=r e^{-r^2/2\sigma^2}$

Also, the term in the last integrals is wrong $- \sigma^2 r e^{-r^2/2\sigma^2} $.

One easy way to look at the results is to write $$\int_0^\infty r^2 e^{-r^2/2\sigma^2} dr=\frac{\sqrt{2\pi}}{\sqrt{2\pi}}\int_0^\infty r^2 e^{-r^2/2\sigma^2} dr$$ and use the change of variable $u=\frac{r}{\sigma}$

$$\int_0^\infty r^2 e^{-r^2/2\sigma^2} dr=\sigma^3\sqrt{2\pi}\frac{1}{\sqrt{2\pi}}\int_0^\infty u^2 e^{-u^2/2} dr$$

We know that if we define $X$ a standard normal variable(variance equals $1$), $E(X^2 1_{X>0})=\frac{1}{\sqrt{2\pi}}\int_0^\infty u^2 e^{-u^2/2} $ , which is also equal to $\frac{1}{2}$, by symmetry.

The final result is then $$\frac{\sigma^3\sqrt{2\pi}}{2}$$


$$\int_0^\infty r^2 e^{-r^2/2\sigma^2} dr$$

Let $r=\sigma\cdot x$, then we have:

$$\int_0^\infty \sigma^2x^2 e^{-x^2/2} \sigma dx=\sigma^3 \int_0^\infty x^2 e^{-x^2/2}dx$$

Now, we appeal to integration by parts:

$$\int_0^\infty x^2 e^{-x^2/2}dx=\int_0^\infty \left(-e^{-x^2/2}\right)'(x)dx$$

$$=-xe^{-x^2/2}|_0^\infty+\int_0^\infty e^{-x^2/2}\,dx=(0-0)+\sqrt{\frac{\pi}{2}}$$

by the celebrated Gaussian integral.

This gives the final solution as $\sigma^3 \sqrt{\frac{\pi}{2}}$.


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