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Triple Integral

  1. Aug 3, 2005 #1
    I need to prove the following:

    [tex]\int_{0}^1\int_{0}^1\int_{0}^1\frac{1}{1-xyz}dxdydz=\sum_{n=1}^{\infty}\frac{1}{n^3}[/tex]

    Or, as a generalization:

    [tex]\int_{0}^1\cdots\int_{0}^1\frac{1}{1-\prod_{k=1}^mx_k}\prod_{k=1}^mdx_k=\sum_{n=1}^{\infty}\frac{1}{n^m}[/tex]

    ...if there is such a generalization.

    I don't know where to begin, any suggestions?

    Thanks a lot for your help.
     
    Last edited: Aug 3, 2005
  2. jcsd
  3. Aug 3, 2005 #2
    You can make a Taylor expansion of the denominator since it is convergent for every point in the domain of integration.
     
  4. Aug 4, 2005 #3
    Alright, a friend showed me how to do this for m=3. Does this work?

    [tex]\sum_{n=1}^{\infty}\frac{1}{n^3}=\sum_{n=0}^{\infty}\frac{1}{(n+1)^3}[/tex]

    and since [tex]\int_{0}^1x^kdx=\frac{1}{k+1}[/tex], the sum can be rewritten as follows:

    [tex]\sum_{n=0}^{\infty}\frac{1}{(n+1)^3}=\sum_{n=0}^{\infty}\int_{0}^1\int_{0}^1\int_{0}^1(xyz)^kdxdydz=\int_{0}^1\int_{0}^1\int_{0}^1\frac{1}{1-xyz}dxdydz[/tex].

    This works? Or, is the proof more in-depth? I would like to know whether the needs to be a justification for evaluating the outside sum first.

    Thanks again.
     
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