How to evaluate this double integeration of a gaussian function?

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The discussion focuses on the integration of the double integral of a Gaussian function represented by the equation ∫∫e^-x^2 e^-x'^2 / |x-x'| dx^3 dx'^3, with limits from 0 to infinity for both variables. The integration process leads to the error function, and participants suggest using polar coordinates to simplify the evaluation. Specifically, the transformation to polar coordinates, where x = r sin(θ) and x' = r cos(θ), is recommended to facilitate the integration of the product of two Gaussian functions.

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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
 
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Perhaps try polar coordinates.
x=r sin /theta
x' = r cos /theta
 
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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