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

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

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} $.