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 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. $$\sum_{\rho} \frac{1}{\rho (1{-}\rho)}=\sum_{\rho} \frac{1}{|\rho|^2} = 2 + \gamma – \log 4\pi$$.
And the constants still don't match.
Best Answer
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 $$\sum_{ \rho } \frac{1}{\rho(1-\rho)} = 2 \sum_{\Im \rho >0} \frac{1}{\rho(1-\rho)}= 2 +\gamma - \log(4\pi).$$