Originally Posted by

**CaptainBlack** There is a well known trick to evaluating this last integral, which is to consider:

$\displaystyle I^2=\left[\int_0^{\infty}e^{-u^2}du\right]^2=\int_0^{\infty}e^{-x^2}dx \int_0^{\infty}e^{-y^2}dy=\int_{\mathbb{R}^2}e^{-(x^2+y^2)}dxdy$

Now we can rewrite the last integral in polars instead of cartesians:

$\displaystyle \int_{\mathbb{R}^2}e^{-(x^2+y^2)}dxdy=\int_{r=0}^{\infty}\int_{\theta=0}^ {2\pi}e^{-r^2} rd\theta dr$

which can be evaluated by elementary means.

CB