Gaussian integral

<math>\frac{d}{dx} \sin x=?\,</math>

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

The Gaussian integral is the integral:

<math> \int_{-\infty}^{\infty} e^{-x^2}dx. </math>

It has a value of <math>\sqrt\pi</math>. The value is needed to normalize the Normal distribution.

Derivation

First look at the double integral

<math>\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2}dxdy.</math>

Separating it,

<math>\int_{-\infty}^{\infty} e^{-x^2}dx \int_{-\infty}^{\infty} e^{-y^2}dy = (\int_{-\infty}^{\infty} e^{-x^2}dx)^2 </math>

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

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

<math>\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2}dxdy = \int_{0}^{2\pi}\int_{0}^{\infty} e^{-r^2}rdrd\theta</math>

<math> = (\int_{0}^{2\pi}d\theta)(\int_{0}^{\infty} e^{-r^2}rdr)</math>

<math> = 2\pi \int_{0}^{\infty} e^{-r^2}rdr.</math>

Substituting <math>z=r^2</math> into the integral:

<math> = 2\pi \int_{0}^{\infty} e^{-z}(z^{1/2})(\frac{dz}{2z^{1/2}})</math>

<math> = \pi \int_{0}^{\infty} e^{-z}dz = \pi \left[-e^{-z} \right]_0^{\infty} = \pi.</math>

Therefore,

<math>(\int_{-\infty}^{\infty} e^{-x^2}dx)^2 = \pi </math>

<math>\int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt\pi.</math>