I am trying to find or get a numerical approximation of
$$ \sum\_{\rho \text{ non-trivial zeros of } \zeta} \frac{1}{\rho} $$


In [The Riemann Hypothesis: Arithmetic and Geometry](http://www.math.lsa.umich.edu/~lagarias/doc/mt-holyoke-rev.pdf) Lagarias gives the identity:


$$\hat{\zeta}(s) := \pi^{-\frac{s}{2}} \Gamma(\frac{s}{2})\zeta(s)$$


$$ \frac{\hat{\zeta}^\prime(s)} {\hat{\zeta}(s)} = \frac{d}{ds} [ \log \hat{\zeta}(s) ] = -\frac{1}{s} - \frac{1}{s-1} + {\sum\_{\rho \text{ zeros of } \zeta }}^\prime \frac{1}{s-\rho} \qquad(1)$$


where the prime indicates the zeros must be summed in pairs $\rho,1-\rho$



> 
> Q1 Does the last sentence mean that the sum is over the non-trivial zeros?
> 
> 
> 


Maple gives:
$$\lim\_{s \to 0} {\sum\_{\rho \text{ zeros of } \zeta }}^\prime \frac{1}{s-\rho} = -\gamma + \frac{1}{2} \log\left(\pi\right) + \log\left(2\right) - 1$$

If the above result is correct, is it true that:


$$ \sum\_{\rho \text{ non-trivial zeros of } \zeta} \frac{1}{\rho} = \gamma - \frac{1}{2} \log\left(\pi\right) - \log\left(2\right) + 1 $$

**EDIT** As Micah Milinovich kindly answerd the above is wrong.


Trying to save the quiestion, is it true that:
$$ \sum\_{\rho} \frac{1}{\rho (1{-}\rho)} = \gamma - \frac{1}{2} \log\left(\pi\right) - \log\left(2\right) + 1 $$


Assuming RH $1-\rho = \bar{\rho}$ and the LHS is $\sum\_{\rho} \frac{1}{|\rho|^2}$


According to [RH Equivalence 5.3](http://aimath.org/pl/rhequivalences). $$\sum\_{\rho} \frac{1}{\rho (1{-}\rho)}=\sum\_{\rho} \frac{1}{|\rho|^2} = 2 + \gamma - \log 4\pi$$.


And the constants still don't match.




Is this sum of reciprocals of zeta zeros correct?
To answer your modified question, according to Mathematica:


$$ \lim\_{s\to 0} \left(\frac{\hat{\zeta}'}{\hat{\zeta}}(s)+\frac{1}{s}\right) = -\frac{\gamma}{2} + \tfrac{1}{2}\log(4\pi).$$


This implies that


$${\sum\_\rho}'\frac{1}{\rho} = \sum\_{\Im \rho >0} \frac{1}{\rho(1-\rho)}= 1 +\frac{\gamma}{2} - \tfrac{1}{2}\log(4\pi). $$


Therefore
$$\sum\_{ \rho } \frac{1}{\rho(1-\rho)} = 2 \sum\_{\Im \rho >0} \frac{1}{\rho(1-\rho)}= 2 +\gamma - \log(4\pi).$$


