Statement

For a>0a>0, the Gaussian integral is

Reax2dx=π/a.\int_{\mathbb R}e^{-ax^2}\,dx=\sqrt{\pi/a}.

Its product form is Rneax2dx=(π/a)n/2\int_{\mathbb R^n}e^{-a|x|^2}\,dx=(\pi/a)^{n/2}, by .

Planar proof

The integral for a=1a=1 is finite and positive. Its square equals R2e(x2+y2)dxdy\int_{\mathbb R^2}e^{-(x^2+y^2)}\,dx\,dy. In polar coordinates this becomes 2π0er2rdr=π2\pi\int_0^\infty e^{-r^2}r\,dr=\pi. Taking the positive square root gives π\sqrt\pi; substitution y=axy=\sqrt a\,x gives the general case.