The Gaussian integral is the integral:
It has a value of Ï {\displaystyle {\sqrt {\pi }}} . The value is needed to normalize the Normal distribution.
First look at the double integral
â« â â â â« â â â e â x 2 â y 2 d x d y . {\displaystyle \int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{-x^{2}-y^{2}}dxdy.}
Separating it,
â« â â â e â x 2 d x â« â â â e â y 2 d y = ( â« â â â e â x 2 d x ) 2 {\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}dx\int _{-\infty }^{\infty }e^{-y^{2}}dy=(\int _{-\infty }^{\infty }e^{-x^{2}}dx)^{2}}
So, the double integral is merely the square of the Gaussian integral.
Now, do the double integral in polar co-ordinates. â x 2 â y 2 = â ( x 2 + y 2 ) = â r 2 {\displaystyle -x^{2}-y^{2}=-(x^{2}+y^{2})=-r^{2}} and d x d y = r d r d θ {\displaystyle dxdy=rdrd\theta } , so:
â« â â â â« â â â e â x 2 â y 2 d x d y = â« 0 2 Ï â« 0 â e â r 2 r d r d θ {\displaystyle \int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{-x^{2}-y^{2}}dxdy=\int _{0}^{2\pi }\int _{0}^{\infty }e^{-r^{2}}rdrd\theta }
= ( â« 0 2 Ï d θ ) ( â« 0 â e â r 2 r d r ) {\displaystyle =(\int _{0}^{2\pi }d\theta )(\int _{0}^{\infty }e^{-r^{2}}rdr)}
= 2 Ï â« 0 â e â r 2 r d r . {\displaystyle =2\pi \int _{0}^{\infty }e^{-r^{2}}rdr.}
Substituting z = r 2 {\displaystyle z=r^{2}} into the integral,
= 2 Ï â« 0 â e â z ( z 1 / 2 ) ( d z 2 z 1 / 2 ) {\displaystyle =2\pi \int _{0}^{\infty }e^{-z}(z^{1/2})({\frac {dz}{2z^{1/2}}})}
= Ï â« 0 â e â z d z = Ï [ â e â z ] 0 â = Ï . {\displaystyle =\pi \int _{0}^{\infty }e^{-z}dz=\pi \left[-e^{-z}\right]_{0}^{\infty }=\pi .}
Therefore,
( â« â â â e â x 2 d x ) 2 = Ï {\displaystyle (\int _{-\infty }^{\infty }e^{-x^{2}}dx)^{2}=\pi }
â« â â â e â x 2 d x = Ï . {\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}dx={\sqrt {\pi }}.}
and 
, so:
into the integral,
![{\displaystyle =\pi \int _{0}^{\infty }e^{-z}dz=\pi \left[-e^{-z}\right]_{0}^{\infty }=\pi .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d5336cdbf12de9441ca5a5914951eb17aa7a128b)
