Sum of Series Zeros Riemann function It is posible to prove the equality  of the following series
$$-\sum _{k=1}^{\infty } \left(\frac{2^{2 k+1} \pi ^{2 k}}{4 k^2 (k+2) (-1)^k (2 k)! \zeta (2 k+1)}+\frac{4}{2 \left(\rho _k-4\right) \left(\rho _k\right){}^2 \zeta '\left(\rho _k\right)}\right)+\frac{1}{4}+\frac{45}{4 \pi ^4}=\log (2 \pi )$$ Numerically for a few Zeros it seem to work
 A: Iff all the zeros are simple then the residue theorem gives
$$0=\lim_{\sigma\to \infty} \frac{1}{2i\pi}\int_{\sigma-i\infty}^{\sigma+i\infty} \frac{1}{\zeta(s)} \frac1{(s-4)s^2}ds=\frac{1}{2i\pi}\int_{5-i\infty}^{5+i\infty} \frac{1}{\zeta(s)} \frac1{(s-4)s^2}ds$$ $$ = 
Res_{s=0}(\frac{1}{\zeta(s)} \frac1{(s-4)s^2})+Res_{s=4}(\frac{1}{\zeta(s)} \frac1{(s-4)s^2})$$
$$+\sum_\rho Res_{s=\rho}(\frac{1}{\zeta(s)} \frac1{(s-4)s^2})+\sum_{k=1}^\infty Res_{s=-2k}(\frac{1}{\zeta(s)} \frac1{(s-4)s^2})$$
$$ = \frac{1}{\zeta(0)(-4)}-\frac{1}{\zeta(0)(-4)^2}-\frac{\zeta'(0)}{\zeta(0)^2(-4)}+\frac{1}{\zeta(4) 4^2}$$
$$ + \sum_\rho \frac{1}{\zeta'(\rho)} \frac1{(\rho-4)\rho^2}+\sum_{k=1}^\infty \frac{1}{\zeta'(-2k)} \frac1{(-2k-4)(-2k)^2}$$
Proving the convergence of the series over the non-trivial zeros needs a lot of estimates (density of zeros, Phragmén–Lindelöf, Hadamard 3 circles...)
The functional equation is $\zeta(s) = \chi(s) \zeta(1-s)$ with the closed-form for $\chi'(s)$ we have $$\zeta'(-2k) = \chi'(-2k)\zeta(2k+1)= (-1)^k \frac {(2k)!} {2 (2\pi)^{2k}} \zeta (2k+1)$$
$$\frac{\zeta'(0)}{\zeta(0)} = \log (2\pi)$$
