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How to show integral identity involving gaussian over x

  1. Jan 25, 2012 #1
    So I have come across this integral identity in Krall's Principles of Plasma Physics (right after equ. 6.4.4 in the 1st edition) and I have not been able to show the identity is true. The reason that I would like to understand the integral is that I am trying to solve a similar problem to the one done in the book. I am hoping that understanding how solve the integral given below will help me with others.

    The identity that Krall uses is:

    [itex]\int d\vec{x} \frac{ e^{- x^2 }}{ \left| \vec{x}-\vec{y} \right| } = \frac{\pi^{3/2}\text{Erf}\left[ y \right]}{y} [/itex]

    Where x,y denote the magnitude of the x,y vectors and Erf is the error function. The integral on x is over all 3D space.

    I suppose in the end I want to know how to show the identity is true. It would also be helpful is someone has seen this or similar integrals in table of integrals. Resources that are related are welcomed. I would also just like some more ideas on how to attack it.

    My best attempts were using:
    spherical coordinates (with z-axis along y-direction)
    differentiating under the integral with respect to magnitude if y

    I'm in my 3rd year of grad. school in physics so hopefully you won't need to spell it all out for me to follow. Thanks!
     
  2. jcsd
  3. Jan 26, 2012 #2
    I solved it (after some friends pointed out some simple mistakes I was making).


    For the record, here is how I did it. I will use spherical coordinates with the z-axis aligned with [itex]\vec{y}[/itex]

    [itex] \int d\vec{x} \frac{e^{-x^2}}{\left| \vec{x} - \vec{y}\right|} = \int_0^{2 \pi} d\theta \int_0^\pi d\phi \int_0^\infty dr \frac{r^2 \sin \phi e^{-r^2}}{\sqrt{r^2 + y^2 - 2 y r \cos \phi}}[/itex]

    Now do the [itex]\theta,\phi[/itex] integrals


    [itex] \int d\vec{x} \frac{e^{-x^2}}{\left| \vec{x} - \vec{y}\right|} = \int_0^\infty dr \frac{2\pi r}{y} e^{-r^2} \left( r + y - \sqrt{(r-y)^2} \right)[/itex]

    The cute thing to notice (pointed out by my friend) is that we must treat [itex] \sqrt{(r-y)^2}[/itex] in parts for when [itex]r>y[/itex] and [itex]y>r[/itex]. Now we have two integrals:


    [itex]\frac{4\pi}{y}\int_0^y dr r^2 e^{-r^2} + 4\pi \int_y^\infty dr r e^{-r^2}[/itex]

    Then we can just look these integrals up in a table or use Mathematica. I hope this helps someone who gets stuck on this integral like me.
     
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