# Would be this formula valid ? Zeta regularization and Euler product plus zeros.

$$\frac{\zeta (0)}{\zeta (s)}= \prod _{p}\prod_{m= -\infty}^{\infty}\left(1-\frac{is\log p}{2\pi m}\right),$$

where $m$ does not run over $m=0$ and '$p$' means a product over all the primes :)

Is this valid ? I have used the Euler product representation

$$\frac{1}{ \zeta (s)}= \prod _{p}(1-p^{-s}).$$

The zeros of $1-p^{-s}$ , are given by $\frac{2\pi i m}{ \log(p)}$ so I think this would be valid , here $i= \sqrt{-1}$.

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There are serious convergence issues. – blabler Jun 26 '13 at 22:48
No. $$\prod_{m=1}^{\infty}\left(1-\frac{x^2}{m^2}\right)=\frac{\sin\pi x}{\pi x}.$$ – Start wearing purple Jun 26 '13 at 22:48
@OL the product over m gives your fucntion but we mus also include a product over primes – Jose Garcia Jun 26 '13 at 22:51
@JoseGarcia For the product to make sense (not to speak about true convergence) the factors should tend to $1$ as $p\rightarrow \infty$. – Start wearing purple Jun 26 '13 at 22:53
@OL: There is a problem with the convergence of the sines. If not then please provide the region of convergence. – blabler Jun 26 '13 at 23:17