How to prove

∫∞0e−x2dx=√π2

**Answer**

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.

**Attribution***Source : Link , Question Author : Jichao , Answer Author : Ross Millikan*