# Difference between revisions of "Gaussian integral"

From Conservapedia

(drop the quotation marks!) |
(Seperatint -> Separating, Substitutiing -> Substituting) |
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<math>\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2}dxdy.</math> | <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> | <math>\int_{-\infty}^{\infty} e^{-x^2}dx \int_{-\infty}^{\infty} e^{-y^2}dy = (\int_{-\infty}^{\infty} e^{-x^2}dx)^2 </math> | ||

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

## Revision as of 06:47, 18 September 2011

- It has been proposed that this page,
**Gaussian integral**, be titled, "".**Gaussian integral**

(The current version uses quotation marks in the title!)

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

The **Gaussian integral** is the integral:

It has a value of . The value is needed to normalize the Normal distribution.

## Derivation

First look at the double integral

Separating it,

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

Now, do the double integral in polar co-ordinates. and , so:

Substituting into the integral,

Therefore,