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In the wikipedia the Hadamard product for the Riemann's zeta function has two forms. The first one is $\zeta(s)=\frac{e^{(\log(2\pi)-1-\gamma/2)s}}{2(s-1)\Gamma(1+s/2)}\prod_{\rho}\left(1-\frac{s}{\rho} \right)e^{s/\rho}$

The second one is simplified to the form $\zeta(s)=\frac{\pi^{s/2}}{2(s-1)\Gamma(1+s/2)}\prod_{\rho}\left(1-\frac{s}{\rho} \right)$

So my question is, how to prove the identity $e^{(\log(2\pi)-1-\gamma/2)s}\prod_{\rho}e^{s/\rho}= \pi^{s/2}$

without even know the nontrivial roots, $\rho$, of $\zeta(s)$?


Edit 1:

This is equivalent to calculate the value of $\sum_{\rho}\frac{1}{\rho}$ But, how?

1 Answers 1

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Use the fact that $Z(n)=\sum_k \rho_k^{-n}$ can be calculated explicitely for $n=1$ to: $ Z(1)=\frac12[2+\gamma-\ln(4\pi)]=1+\gamma/2-\log(2\pi^{1/2}),\tag{$\star$} $ eq. (5) from here. Take the log of your equation in question and see that $s$ cancels out: $ {(\log(2\pi)-1-\gamma/2)}+\sum_{\rho}{\rho^{-1}}= \log \pi^{1/2}\\ {(\log(2\pi)-1-\gamma/2)}+(1+\gamma/2-\log(2\pi^{1/2}))=\log{\frac{2\pi}{2\pi^{1/2}}}= \frac12 \log \pi. $ Unfortunately I don't have a proof for $(\star)$ yet...