The Gaussian integral is the integral:
It has a value of π {\displaystyle {\sqrt {\pi }}} . The value is needed to normalize the Normal distribution.
First look at the double integral
∫ − ∞ ∞ ∫ − ∞ ∞ e − x 2 − y 2 d x d y . {\displaystyle \int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{-x^{2}-y^{2}}dxdy.}
Separating it,
∫ − ∞ ∞ e − x 2 d x ∫ − ∞ ∞ e − y 2 d y = ( ∫ − ∞ ∞ e − 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. − x 2 − y 2 = − ( x 2 + y 2 ) = − 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:
∫ − ∞ ∞ ∫ − ∞ ∞ e − x 2 − y 2 d x d y = ∫ 0 2 π ∫ 0 ∞ e − 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 }
= ( ∫ 0 2 π d θ ) ( ∫ 0 ∞ e − r 2 r d r ) {\displaystyle =(\int _{0}^{2\pi }d\theta )(\int _{0}^{\infty }e^{-r^{2}}rdr)}
= 2 π ∫ 0 ∞ e − 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 π ∫ 0 ∞ e − 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}}})}
= π ∫ 0 ∞ e − z d z = π [ − e − z ] 0 ∞ = π . {\displaystyle =\pi \int _{0}^{\infty }e^{-z}dz=\pi \left[-e^{-z}\right]_{0}^{\infty }=\pi .}
Therefore,
( ∫ − ∞ ∞ e − x 2 d x ) 2 = π {\displaystyle (\int _{-\infty }^{\infty }e^{-x^{2}}dx)^{2}=\pi }
∫ − ∞ ∞ e − x 2 d x = π . {\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)
