Is There a Way to Regularize Euler Products on Primes?

[mhill]

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

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

 

[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.<br /> Differentiating then gives,<br /> <br /> $$\frac{\zeta'(s)}{\zeta(s)} = -\sum_{p} (p^{s} - 1)^{-1} \log{p}$$<br /> <br /> 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,<br /> <br /> $$\prod_{n} \lambda_{n} = \exp{-\zeta_{\lambda}'(0)}$$<br /> <br /> for a zeta function defined on a sequence [itex](\lambda_{n})_{n \geq 1}[/tex].<br /> <br /> If its a prime regularization you're after, look for this paper by Munoz Garcia and Perez Marco called 'Super Regularization of Infinite Products'.<br /> <br /> Never mind, here's the link to the preprint pdf-<br /> http://inc.web.ihes.fr/prepub/PREPRINTS/M03/M03-52.pdf[/itex][/itex][/itex]