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How to evaluate this double integeration of a gaussian function?

  1. Sep 20, 2011 #1
    how to integrate this equation?

    ∫∫e^-x^2 e^-x'^2 / |x-x'| dx^3 dx'^3 lower limit is 0 and upper limit is inf for x and x'

    the result is an error function. But I would like to know the details of the process of integration.

    some one suggest me to change the variable to x+x'=u x-x'=v but I got stucked. Can some one gave me any further suggestions or hints. Much appreciated!

    With Best Regards
    Tsung-Wen Yen
  2. jcsd
  3. Sep 20, 2011 #2
    Perhaps try polar coordinates.
    x=r sin /theta
    x' = r cos /theta
  4. Sep 23, 2011 #3
    To extend on Kdblin78's comment:

    Let I:= Int e^(-x^2)

    Consider e^(-x^2) , and e^(-y^2)

    Then consider I^2 as the product of the two integrals, and use polar coordinates

    like Kdbnlin78 suggested, to evaluate. Notice that e^(-x^2) is a constant when

    integrating with respect to y, and viceversa for e^(-y^2) . Then I^2 is the

    integral of e^(-x^2- y^2 ), and polar kicks-in nicely.
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