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Zeta regularization riddle

  1. Aug 22, 2008 #1
    examining the zeta regularization of divergent series i found

    [tex] \int_{0}^{\infty} x^{m}dx = (m/2) \int_{0}^{\infty}x^{m-1}dx +\zeta (-m) - \sum_{r=1}^{\infty} \frac{B_ {2r} \Gamma (m+1) (m-2r+1)}{ (2r)! \Gamma (m-2r+2)} \int_{0}^{\infty} x^{m-2r}dx [/tex]

    of course this is only justified considering we take the sum of the series

    [tex] 1+2^{r} + 3^{r} + ................. = \zeta (-r) [/tex]

    B_2r are just the Bernoulli numbers.

    Also another Brain Teaser...

    Let be R(x) the Riesz function , see http://en.wikipedia.org/wiki/Riesz_criterion then we define [tex] R(x^{2} ) = f(x) [/tex]

    then prove that f(x) satisfy the Integral equation

    [tex] \frac{1-e^{x^2} }{2} = \int_{0}^{\infty} \frac{dt}{t} [x/t] f(t) [/tex] , with [tex] \zeta(s)=s\int_{0}^{\infty} dx [x]x^{-s-1} [/tex]
     
  2. jcsd
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