Integral of $\frac{r^{2}}{(2 \pi \sigma^{2})^{\frac{3}{2}}}\exp\left(-\frac{1}{2}(\frac{r}{\sigma})^{2}\right)$

Question

I'm trying to solve the integral of the following function in a sphere of radius $5\sigma$

the function is: $$f(r) = \frac{r^{2}}{(2 \pi \sigma^{2})^{\frac{3}{2}}}\exp\left(-\frac{1}{2}(\frac{r}{\sigma})^{2}\right)$$

and i have to compute the following integral $2\pi^{2}\int_{0}^{5\sigma}f(r)dr$ (the factor of $2 \pi^{2}$ comes from the integration along $\theta$ and $\phi$ in spherical coordinates)

by solving with mathematematica i get that the result is independent of $\sigma$ in particular i get 1.57077

Moreover ,whatever $\sigma$ i choose, the integral result tends to $\frac{\pi}{2}$ as the upper limit of the integral goes to $\infty$ (With a fixed $\sigma$)

is this correct?

Another remark: i'm also solving this integral with a finite element method (gaussian quadrature) on a cubic mesh. Also using this method the integral is independent from $\sigma$ however as i raise the upper limit of the integral, in this case, the integral result asymptotize to $1$. There's a ratio between the results that is exactly $\frac{\pi}{2}$. Any idea of some possible mistake?

Answer 1

Your integral is "almost" the second moment of the Normal distribution. The only difference is a constant factor $\frac {1} {2\pi \sigma^2}$ in front of the integral times $\frac {1}{2}$ to account for different integration limits. The second moment of the Normal distribution is $\sigma^2$. So if you multiply by $2\pi^2$ your integral, you are supposed to get

$$ 2\pi^2 \frac {1} {4\pi \sigma^2} \sigma^2=\pi/2$$

Take a look at the Wikipedia link below.

http://en.wikipedia.org/wiki/Normal_distribution

Answer 2

a sphere of radius $5\sigma$

From a practical perspective, that's the same as integrating it over R, since the bulk of the error function, whose derivative your integral is, is comprised within a radius of $3\sigma$.

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Let $I(a)=\displaystyle\int_0^\infty e^{-ax^2}~dx$. Can you evaluate it ? Can you also evaluate $I'(a)$, by differentiating under the integral sign with regard to a ? Now let $a=-\dfrac1{2\sigma^2}$.