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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?

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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...

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