Gaussian integral

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

This article/section deals with mathematical concepts appropriate for a student in late high school or early university.

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

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

Separating 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 $d x d y = r d r d θ$ , so:

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

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

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

Substituting $z = r 2$ 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,

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

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