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Line 13:
Line 13: − Seperating it,+ Line 27:
Line 27: − Substitutiing <math>z=r^2</math> into the integral,+
Gaussian integral (view source)
Revision as of 06:47, September 18, 2011
, 06:47, September 18, 2011Seperatint -> Separating, Substitutiing -> Substituting
<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>
<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>