This integral popped up whilst finding the expectancy of the distance when both $x$ and $y$ are Normal $(0,\sigma^2)$ distributed (so independent): $$\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\sqrt{x^2+y^2}e^{-\frac{1}{2\sigma^2}(x^2+y^2)}dydx$$ Now, I tried substituting $y=x\tan\theta$, but that did not make it any easier for me.
How can I solve this?
On
Switching to polar coordinates, we have
$$x=r\cos\theta$$ $$y=r\sin\theta$$
This is helpful because then we have
$$x^2+y^2 = r^2\cos^2\theta + r^2\sin^2\theta = r^2$$
The Jacobian of this transformation is just $r$, so we have:
\begin{align} I=&\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\sqrt{x^2+y^2}e^{-\frac{1}{2\sigma^2}(x^2+y^2)}dydx\\&=\frac{1}{2\pi}\int_0^\infty\int_0^{2\pi}r^2e^{-\frac{r^2}{2\sigma^2}}drd\theta\\ &=\int_0^\infty r^2e^{-\frac{r^2}{2\sigma^2}}dr \end{align}
Letting $u:=\frac{r}{\sqrt{2}\sigma}$, then $r^2=2\sigma^2u$ and $dr=\sqrt{2}\sigma du$. The integral becomes
\begin{align} I&=\sqrt{8}\sigma^3\int_0^\infty u^2e^{-u^2}du\\ &=-\sqrt{8}\sigma^3\int_0^\infty \frac{\partial}{\partial\mu}e^{-\mu u^2}du\bigg\vert_{\mu=1}\\ &=-\sqrt{8}\sigma^3\frac{d}{d\mu}\int_0^\infty e^{-\mu u^2}du\bigg\vert_{\mu=1}\\ &=-\sqrt{2\pi}\sigma^3\frac{d}{d\mu}\left(\frac{1}{\sqrt{\mu}}\right)\bigg\vert_{\mu=1}\\ &=\sqrt{2\pi}\sigma^3\cdot\frac{1}{2\mu^{\frac{3}{2}}}\bigg\vert_{\mu=1}\\ &=\sqrt{\frac{\pi}{2}}\sigma^3 \end{align}
I hope that I didn't make a mistake with the constants.
In polar coordinates, the integral is $$ \frac{1}{2\pi}\int_0^\infty \int_0^{2\pi}r^2e^{-\frac{r^2}{2\sigma^2}}dr d\theta=\int_0^\infty r^2 e^{-\frac{r^2}{2\sigma^2}}dr $$ which by parts is $$ -r\sigma^2e^{-\frac{r^2}{2\sigma^2}}\vert_0^\infty+\sigma^2\int_0^\infty e^{-\frac{r^2}{2\sigma^2}}dr=\sigma^2\int_0^\infty e^{-\frac{r^2}{2\sigma^2}}dr $$ where hopefully I did not screw up with the constants floating around and with the final integral is the gaussian.