Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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} $.

share|improve this question
    
There are serious convergence issues. –  blabler Jun 26 '13 at 22:48
    
No. $$\prod_{m=1}^{\infty}\left(1-\frac{x^2}{m^2}\right)=\frac{\sin\pi x}{\pi x}.$$ –  O.L. 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 $p\rightarrow \infty$. –  O.L. 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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.