Calculating Integrals using Dimensional Regularization

[robousy]

Hey folks,

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

I'm trying to show:


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

I know the expression for the Gamma function is

$$\Gamma(s)=\int_0^\infty t^{s-1}e^{t}dt$$

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

I also know,

$$z^{-s}=\frac{1}{\Gamma(s)}\int_0^\infty t^{s-1}e^{-zt}$$

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!

 

[mhill]

HINT 1: see the volume of 2n-dimensional Sphere at

http://en.wikipedia.org/wiki/Dimensional_regularization

HINT 2: for the integral over 't' $$\int_{0}^{\infty} dx e^{-ux}x^{r} = \Gamma(r+1)u^{-r-1}$$

HINT 3: $$\pi cotg(i \pi z) = (-i/z)-2iz \sum_{n=1}^{\infty}(z^{2}+n^{2})^{-1}$$

cot(x) means cotangent.