- #1
robousy
- 334
- 1
I'm trying to work out the steps in the following:
[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-1}e^{-(k^2+a^2m^2)t}dt=-\frac{\pi^n}{(2\pi)^{2n}L^{2n}}\frac{\Gamma(s-n)}{\Gamma(s)}L^{2s}\zeta(2s-2n)[/tex]
I've tried a number of things including using the Jacobi theta function to change the limits on the summation, integral representation of the modified bessel function, and various expressions for the gamma and zeta functions, but to no avail. I've spent about 3 days on this so far. If anyone experienced in regularization can help me get from the left hand side to the right hand side I will be forever grateful!
Richard
[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-1}e^{-(k^2+a^2m^2)t}dt=-\frac{\pi^n}{(2\pi)^{2n}L^{2n}}\frac{\Gamma(s-n)}{\Gamma(s)}L^{2s}\zeta(2s-2n)[/tex]
I've tried a number of things including using the Jacobi theta function to change the limits on the summation, integral representation of the modified bessel function, and various expressions for the gamma and zeta functions, but to no avail. I've spent about 3 days on this so far. If anyone experienced in regularization can help me get from the left hand side to the right hand side I will be forever grateful!
Richard
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