Proving ∫∞0e−x2dx=π√2

This is an old favorite of mine.
Define

I=∫+∞−∞e−x2dx

Then

I2=(∫+∞−∞e−x2dx)(∫+∞−∞e−y2dy)

I2=∫+∞−∞∫+∞−∞e−(x2+y2)dxdy

Now change to polar coordinates

I2=∫+2π0∫+∞0e−r2rdrdθ

The θ integral just gives 2π, while the r integral succumbs to the substitution u=r2

I2=2π∫+∞0e−udu/2=π

So

I=π√

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.