# Decomposition of Double Sum

1. Feb 15, 2008

### robousy

Hey folks,

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

Has anyone seen this relation? Is it standard?

Thanks!

Last edited: Feb 15, 2008
2. Feb 16, 2008

### Rainbow Child

Are you sure this is the correct relation? Or is it like this:

$$\sum_{m,n=-\infty}^\infty '(n^2+a^2m^2)^{-s}=-\sum_{n=-\infty}^\infty 'n^{-2s}+\sum_{m=-\infty}^\infty \sum_{n=-\infty}^\infty '(n^2+a^2m^2)^{-s}$$

3. Feb 16, 2008

### robousy

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.

Thanks!

Richard

4. Feb 16, 2008

### robousy

aaaah....thats simple actually, so yes you are probably correct, the extra term is just the m=0 term. lol.

5. Feb 16, 2008

### Rainbow Child

I hope that's correct! Or else we are both missing something!

6. Feb 16, 2008

### Pere Callahan

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 ...

7. Feb 16, 2008

### robousy

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.

8. Feb 16, 2008

### Rainbow Child

Loop gravity! Nice subject, although not my favorite one!

Good luck with the calculations, as I can remember from the regularization in QFT, one mistake makes everything look like Chinese! :rofl:

9. Feb 16, 2008