Gaussian integral

${\frac {d}{dx}}\sin x=?\,$

This article/section deals with mathematical concepts appropriate for late high school or early college.

The Gaussian integral is the integral:

$\int _{-\infty }^{\infty }e^{-x^{2}}dx.$

It has a value of ${\sqrt {\pi }}$. The value is needed to normalize the Normal distribution.

Derivation

First look at the double integral

$\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{-x^{2}-y^{2}}dxdy.$

Seperating it,

$\int _{-\infty }^{\infty }e^{-x^{2}}dx\int _{-\infty }^{\infty }e^{-y^{2}}dy=(\int _{-\infty }^{\infty }e^{-x^{2}}dx)^{2}$

So, the double integral is merely the square of the Gaussian integral.

Now, do the double integral in polar co-ordinates. $-x^{2}-y^{2}=-(x^{2}+y^{2})=-r^{2}$ and $dxdy=rdrd\theta$, so:

$\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{-x^{2}-y^{2}}dxdy=\int _{0}^{2\pi }\int _{0}^{\infty }e^{-r^{2}}rdrd\theta$

$=(\int _{0}^{2\pi }d\theta )(\int _{0}^{\infty }e^{-r^{2}}rdr)$

$=2\pi \int _{0}^{\infty }e^{-r^{2}}rdr.$

Substitutiing $z=r^{2}$ into the integral,

$=2\pi \int _{0}^{\infty }e^{-z}(z^{1/2})({\frac {dz}{2z^{1/2}}})$

$=\pi \int _{0}^{\infty }e^{-z}dz=\pi \left[-e^{-z}\right]_{0}^{\infty }=\pi .$

Therefore,

$(\int _{-\infty }^{\infty }e^{-x^{2}}dx)^{2}=\pi$

$\int _{-\infty }^{\infty }e^{-x^{2}}dx={\sqrt {\pi }}.$