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Gaussian integral

  1. Oct 24, 2011 #1
    How to show that the variance of the gaussian distribution using the probability function? I don't know how to solve for ∫r^2 Exp(-2r^2/2c^2) dr .
  2. jcsd
  3. Oct 24, 2011 #2


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    Use integration by parts and a substitution. It's really closely related to the integral of Exp(r^2).
    Last edited: Oct 24, 2011
  4. Oct 24, 2011 #3
    I tried it. The probability function is 1/(sqrt(2Pi c^2)) * Exp[-r^2/2c] When integrate it from -infinity to infinity, the Exp[r^2] makes everything 0. But we are trying to proof that it's equal to c.
  5. Oct 25, 2011 #4

    Ray Vickson

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    Absolutely not: the integral of exp(-x^2) for x going from - infinity to + infinity is a finite, positive value (it is the area under the curve of the graph y = exp(-x^2)); furthermore, this integral can be found everywhere in books and web pages; I will let you find it.

    Anyway, you need to find an integral of the form int_{x=-inf..inf} x^2*exp(-x^2) dx, which is obtained from yours by an appropriate change of variables, etc. Integrate by parts, setting u = x and dv = x*exp(-x^2) dx.

  6. Oct 25, 2011 #5
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