- #1
robousy
- 334
- 1
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:
[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!
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!
Last edited: