# Sum over all inverse zeta nontrivial zeros

Starting from the Hadamard product for the Riemann Zeta Function (assuming the product is taken over matching pairs of zeros) $$\zeta(s)=\frac{e^{(\log(2\pi)-1-\gamma/2)s}}{2(s-1)\Gamma(1+s/2)}\prod_{\rho}\left(1-\frac{s}{\rho} \right)e^{s/\rho}$$ can one derive the exact value of $\sum_{\rho} \frac{1}{\rho}$ to be $$\sum_{\rho} \frac{1}{\rho} = -\log(2\sqrt{\pi})+1+\gamma/2$$ What implications does this have?

• Basically none. Also note the sum doesn't converge absolutely. – Wojowu Jul 25 '18 at 16:49

Let's start with the logarithm of the Hadamard product and take the derivative relatively to $s$ (remembering that $\;\psi(z):=\dfrac d{dz}\log\Gamma(z)\,$ with $\psi$ the digamma function) : \begin{align} \log\zeta(s)&=(\log(2\pi)-1-\gamma/2)s-\log\left(2(s-1)\Gamma(1+s/2)\right)+\sum_{\rho}\log\left(1-\frac{s}{\rho} \right)+\frac s{\rho}\\ \frac{\zeta'(s)}{\zeta(s)}&=\log(2\pi)-1-\gamma/2-\left(\frac 1{s-1}+\frac 12\psi\left(\frac s2+1\right)\right)+\sum_{\rho}\frac 1{s-\rho}+\frac 1{\rho}\\ \frac{\zeta'(s)}{\zeta(s)}+\frac 1{s-1}&=\log(2\pi)-1-\gamma/2-\frac 12\psi\left(\frac s2+1\right)+\sum_{\rho}\frac 1{s-\rho}+\frac 1{\rho}\\ \end{align} taking the limit as $s\to 1$ and remembering that $\;\displaystyle \zeta(s)-\frac 1{s-1}=\gamma+O(s-1)\;$ and $\psi\left(\frac 32\right)=\psi\left(\frac 12\right)+2=-\gamma-\log(4)+2\;$ we get : \begin{align} \gamma&=\log(2\pi)-1-\gamma/2+\left(\gamma/2+\log(2)-1\right)+\sum_{\rho}\frac 1{1-\rho}+\frac 1{\rho}\\ \sum_{\rho}\frac 1{1-\rho}+\frac 1{\rho}&=-\log(4\pi)+2+\gamma \end{align} With the answer the double of your series (as it should since if $\rho$ is a root then $1-\rho$ will also be one).
For the sum of integer powers of the roots see Mathworld starting at $(5)$.