# integrating $A^2=\frac{1}{2\pi}\int^\infty_{-\infty}\int^\infty_{-\infty}e^{-\frac{y^2+z^2}{2}}dydz$

When proving that $$\int^{\infty}_{-\infty}\frac{1}{\sqrt{2\pi}\sigma}{e^{-\frac{1}{2}({\frac{x-\mu}{\sigma})}^2}}dx=1$$

and I faced a problem, $$A^2=\frac{1}{2\pi}\int^\infty_{-\infty}\int^\infty_{-\infty}e^{-\frac{y^2+z^2}{2}}dydz$$is $$\frac{1}{2\pi}\int^{2\pi}_{0}\int^\infty_{0}e^{-\frac{r^2}{2}}rdrd{\theta}$$

by putting $y=r\sin{\theta},z=r\cos{\theta}$

• Involving the Erf function, the integrals are quite easy to compute in cartesian coordinates. This is also true for the antiderivative. – Claude Leibovici Dec 17 '13 at 10:35
• You are almost done. Put $u=r^2/2$ and work out the problem. – Mhenni Benghorbal Dec 17 '13 at 10:42

You have to change the variables because integration in the original variables is very difficult (i'm not sure how to do it), but changing to polar removes a variable because of the identity $\sin^2{\theta}+\cos^2{\theta}=1$, which you did.
Also, because of the change of variable, you have to calculate the Jacobian which produces an additional term in the integral, namely $r$. Now to continue further, you can just do substition and say, let $u=r^2$ so that $\frac{du}{dr}=2r \rightarrow \frac{du}{2}=rdr$.
Another approach is to write the double integral as $$\frac{4}{2\pi}\int_{0}^{\infty} e^{-x^2/2}d x\int_{0}^{\infty} e^{-y^2/2}d y.$$ Now, all you need to do is to make the same change of variable $u=x^2/2$ and $v=y^2/2$ and then use the gamma function.