# Euler product regularization

1. Oct 28, 2008

### mhill

although is not valid in general (since an Euler product usually converges only when Re(s) >1)

$$\frac{ d \zeta(1/2)}{\zeta (1/2)}= -\sum_{p} log(p)(1-p^{1/2}$$

2. Nov 2, 2008

### yasiru89

Well, this is simply by taking logarithms on either side of the Euler product representation to get,

$$\log{\zeta(s)} = -\sum_{p} \log{(1 - p^{-s})}$$

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

$$\frac{\zeta'(s)}{\zeta(s)} = -\sum_{p} (p^{s} - 1)^{-1} \log{p}$$

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,

$$\prod_{n} \lambda_{n} = \exp{-\zeta_{\lambda}'(0)}$$

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-
http://inc.web.ihes.fr/prepub/PREPRINTS/M03/M03-52.pdf

Last edited: Nov 2, 2008