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Reverted edits by CalebMM (talk) to last revision by Karajou
{{move|Gaussian integral}} (The current version uses quotation marks in the title!)
{{Template:Math-h}}
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|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>
[[category:mathematics]]