# Proving ∫∞0e−x2dx=√π2\int_{0}^{\infty} \mathrm{e}^{-x^2} dx = \frac{\sqrt \pi}{2}

How to prove

This is an old favorite of mine.
Define
Then

Now change to polar coordinates

The $\theta$ integral just gives $2\pi$, while the $r$ integral succumbs to the substitution $u=r^2$

So and your integral is half this by symmetry

I have always wondered if somebody found it this way, or did it first using complex variables and noticed this would work.