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Homework Help: Integral involved in solving Poisson's equation

  1. Feb 1, 2010 #1
    1. The problem statement, all variables and given/known data
    Problem 8.30 from Greenberg's Foundations of Applied Mathematics: We meet the integral

    [tex] I = -\frac{1}{8\pi^3}\int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty\frac{e^{i(\xi x+\eta y+\zeta z)}}{\xi^2+\eta^2+\zeta^2}\,d\xi\,d\eta\,d\zeta [/tex]

    when we solve the Poisson equation by the Fourier transform. Show that [tex] I=-1/4\pi r [/tex], where [tex] r=\sqrt{x^2+y^2+z^2} [/tex]


    2. Relevant equations
    A hint is given; it says to note that [tex] \exp i(\xi x+\eta y+\zeta z)=\exp i\mathbf{R}\cdot\mathbf{r} [/tex] where R is the vector to the point [tex] (\xi ,\eta ,\zeta ) [/tex], r is the vector to the point (x,y,z), and [tex] \theta [/tex] is the angle between R and r. Then change over to spherical polars [tex] R, \theta, \phi [/tex] with r as the polar axis.


    3. The attempt at a solution
    Following the hint, I get

    [tex] I = \int_0^{2\pi}\int_0^\pi\int_0^\infty\frac{e^{irR\cos\theta}}{R^2}R^2\sin\theta\,dR\,d\theta\,d\phi [/tex]

    since the axes are only rotated so the Jacobian is 1 (volumes are not contracted or expanded). However, then I get that the integral doesn't exist. Now I found

    https://www.physicsforums.com/showthread.php?t=293550"

    where apparently a constant squared is thrown into the denominator, integration methods from complex variables are used, and then the limit is taken as the constant approaches zero. However, our class hasn't done complex variables and in any case I am not yet familiar with such methods.

    So have I done something wrong or is there another way to proceed?

    Thanks in advance.
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Feb 2, 2010 #2
    Actually... it does:

    [tex] I = 2\pi\int_0^\infty\,dR\int_0^\pi\sin\theta e^{irR\cos\theta}\,d\theta =2\pi\int_0^\infty\,dR\int_0^\pi (-\frac{1}{irR})\,de^{irR\cos\theta} [/tex]

    [tex] =\frac{4\pi}{r}\int_0^\infty\,dR\frac{1}{R}\frac{e^{irR}-e^{-irR}}{2i}=\frac{4\pi}{r}\int_0^\infty\frac{\sin rR}{R}\,dR=\frac{2\pi^2}{r} [/tex]
     
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