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I've been stuck on this for two days now so I'm hoping for some hints from anyone...

I'm trying to show:

[tex]-\frac{1}{2}\int\frac{d^{2n}k}{(2\pi)^{2n}}\frac{1}{\Gamma(s)}\sum_{m=-\infty}^{m=\infty}\int_0^\infty t^{s-\frac{3}{2}}e^{-(k^2+a^2m^2)t}=-\frac{\pi^n}{(2\pi)^{2n}L^{2n}}\frac{\Gamma(s-n)}{\Gamma(s)}L^{2s}\zeta(2s-2n)[/tex]

I know the expression for the Gamma function is

[tex]\Gamma(s)=\int_0^\infty t^{s-1}e^{t}dt[/tex]

and probably comes in useful somewhere, but I'm not sure where.

I also know,

[tex]z^{-s}=\frac{1}{\Gamma(s)}\int_0^\infty t^{s-1}e^{-zt}[/tex]

which might also come in useful.

I don't know if anyone has much experience with this sort of thing but if you have any tips I'd be grateful!!

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# Dimensional Regularization

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