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A 3-dimensional calculus

  1. Feb 25, 2009 #1
    I'm a researcher of telecommunications. I was stumped by calculation of an integral when doing some proof. Would anyone have any idea how to give a close form of the following 3-dimensional integral? Your help will be highly appreciated! Here, A, B, C are all positive constants.

    [tex]\int_0^{2\pi}\int_0^{2\pi}\int_0^{2\pi}\exp\left(A\cos(y-x)\right)\exp\left(B\cos(z-y)\right)\exp\left(C\cos(z-x)\right)dxdydz.[/tex]

    BTW, I've tried the 2-dimensional case before which is converted to the Modified Bessel function as follow,

    [tex]\int_0^{2\pi}\int_0^{2\pi}\exp\left(A\cos(x-y)\right)dxdy=(2\pi)^2I_0\{A\}[/tex]

    where

    [tex]I_0\{A\}=\frac{1}{2\pi}\int_0^{2\pi}\exp(A\cos\theta)d\theta.[/tex]
     
    Last edited: Feb 25, 2009
  2. jcsd
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