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Dimensional Regularization

  1. Feb 14, 2008 #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!!
    Last edited: Feb 14, 2008
  2. jcsd
  3. Aug 23, 2008 #2
    HINT 1: see the volume of 2n-dimensional Sphere at


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

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

    cot(x) means cotangent.
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