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Decomposition of Double Sum

  1. Feb 15, 2008 #1
    Hey folks,

    I'm trying to show that [tex]\sum_{m,n=-\infty}^\infty '(n^2+a^2m^2)^{-s}=\sum_{n=-\infty}^\infty 'n^{-2s}+\sum_{m=-\infty}^\infty'\sum_{n=-\infty}^\infty(n^2+a^2m^2)^{-s}[/tex]. The prime means that we don't include the m=n=0 term.

    Has anyone seen this relation? Is it standard?

    Last edited: Feb 15, 2008
  2. jcsd
  3. Feb 16, 2008 #2
    Are you sure this is the correct relation? Or is it like this:

    [tex]\sum_{m,n=-\infty}^\infty '(n^2+a^2m^2)^{-s}=-\sum_{n=-\infty}^\infty 'n^{-2s}+\sum_{m=-\infty}^\infty \sum_{n=-\infty}^\infty '(n^2+a^2m^2)^{-s}[/tex]
  4. Feb 16, 2008 #3
    Hey Rainbow,

    Its from a paper so its possible there was a typo in the preprint. Do you know the steps for your version? That would probably help me.


  5. Feb 16, 2008 #4
    aaaah....thats simple actually, so yes you are probably correct, the extra term is just the m=0 term. lol.
  6. Feb 16, 2008 #5
    I hope that's correct! Or else we are both missing something! :smile:
    What was the paper about?
  7. Feb 16, 2008 #6
    It's just splitting the m-sum into m=0 and m != 0, for the m=0 part you have to sum over all inteegers n != 0, if m != 0 there is no such restriction on the value of n ...
  8. Feb 16, 2008 #7
    Its a dimensional regularization process. Are you familiar with calculations of one loop vacuum energy in extra dimensions? Its a sweet little paper on how to do it in a T^2/Z^2 orbifold. I've been working on it all week and its all clicking into place nicely.
  9. Feb 16, 2008 #8
    Loop gravity! Nice subject, although not my favorite one! :smile:

    Good luck with the calculations, as I can remember from the regularization in QFT, one mistake makes everything look like Chinese! :rofl:
  10. Feb 16, 2008 #9
    Your not wrong there!

    Thanks again for the brain jolt.

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