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Euler product regularization

  1. Oct 28, 2008 #1
    although is not valid in general (since an Euler product usually converges only when Re(s) >1)

    [tex] \frac{ d \zeta(1/2)}{\zeta (1/2)}= -\sum_{p} log(p)(1-p^{1/2} [/tex]
  2. jcsd
  3. Nov 2, 2008 #2
    Well, this is simply by taking logarithms on either side of the Euler product representation to get,

    [tex]\log{\zeta(s)} = -\sum_{p} \log{(1 - p^{-s})} [/tex]

    where [itex]p[/tex] is the set of primes.
    Differentiating then gives,

    [tex]\frac{\zeta'(s)}{\zeta(s)} = -\sum_{p} (p^{s} - 1)^{-1} \log{p} [/tex]

    This gives the zeta-regularized sum (and hence product) on primes (which looks curious as it is, unless special values of [itex]s[/tex] are used), but generally its more convenient to consider,

    [tex]\prod_{n} \lambda_{n} = \exp{-\zeta_{\lambda}'(0)}[/tex]

    for a zeta function defined on a sequence [itex](\lambda_{n})_{n \geq 1}[/tex].

    If its a prime regularization you're after, look for this paper by Munoz Garcia and Perez Marco called 'Super Regularization of Infinite Products'.

    Never mind, here's the link to the preprint pdf-
    Last edited: Nov 2, 2008
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