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

Can I have some additional explanations about this?

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    $\begingroup$ Involving the Erf function, the integrals are quite easy to compute in cartesian coordinates. This is also true for the antiderivative. $\endgroup$ – Claude Leibovici Dec 17 '13 at 10:35
  • $\begingroup$ You are almost done. Put $u=r^2/2$ and work out the problem. $\endgroup$ – Mhenni Benghorbal Dec 17 '13 at 10:42
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The integral is much easier to calculate, if the cartesian coordinates are transformed in polar coordinates. A point is given there by the radius r (distance from the point 0/0) and the angle phi. The resulting double integral quickly brings the desired result.

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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 $\frac{du}{dr}=2r \rightarrow \frac{du}{2}=rdr$.

Can you finish the problem from here?

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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.

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