Difference between revisions of "Gaussian integral"

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(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>
  
Seperating it,
+
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>
  
Substitutiing <math>z=r^2</math> into the integral,
+
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,