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given the series $ \sum_{\rho} \frac{1}{z-\rho} $ here the sum is taken OVER the roots of the Riemann function on the critical line $ 0 < Re(s) <1 $

the summation is understood as we sum the pair of zeros $ f( \rho) +f(1-\rho ) $ so $ Re (\rho) > 0 $

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1 Answer 1

up vote 8 down vote accepted

You may start with the Hadamard product : $$ \zeta(s)=\frac{e^{s\left(\log(2\pi)-1-\frac{\gamma}2\right)}}{2(s-1)\Gamma\left(1+\frac s2\right)}\prod_{\rho} \left(1-\frac s{\rho}\right) e^{s/{\rho}} $$ or use the Riemann Xi function $\xi(s)=\frac 12 s(s-1)\pi^{-s/2}\Gamma\left(\frac s2\right)\zeta(s)$ to rewrite this as a $$\xi(s)=\xi(0)\prod_{\rho} 1-\frac s{\rho}$$

so that $\displaystyle \dfrac{\xi'(s)}{\xi(s)} = \sum^\prime_{\rho} \left(\ln\left(1-\frac s{\rho}\right)\right)' = \sum^\prime_{\rho} \frac 1{s-\rho}$
at least if I did'nt make a stupid mistake... (the ' over the sum meaning your summation convention)


You also may find the following derivation interesting: $$\xi(s)=\xi(0)e^{B s}\prod_{\rho} \left(1-\frac s{\rho}\right) e^{s/{\rho}}$$ with $B=\log 2+\frac 12 \log\pi-1-\frac{\gamma}2$ and $\rho$ the non-trivial zeros.

so that $\displaystyle \dfrac{\xi'(s)}{\xi(s)} = B+\sum_{\rho} \left(\ln\left(1-\frac s{\rho}\right)e^{s/{\rho}}\right)' = B+ \sum^\prime_{\rho} \frac 1{s-\rho} + \frac 1{\rho}$

There is no contradiction thanks to $\displaystyle B=-\sum^\prime_{\rho} \frac 1{\rho}=-\sum_{\Im(\rho)>0} \frac 1{\rho(1-\rho)}$

For more see Edwards' book,
Coffey's 'Toward verification of the Riemann hypothesis: Application of the Li criterion'
the MO article 'Is this sum of reciprocals of zeta zeros correct?'
and Lagarias' paper 'The Riemann Hypothesis: Arithmetic and Geometry'

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thanks all the people, who answered, of course if i put $ 1/2+iz $ with 'z' complex i can evaluate the sum only over the imaginary parts $ \sum_{n} (is-it_{n})^{-1} $ –  Jose Garcia Apr 14 '12 at 19:46
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I see only one answer (maybe written by more than one person, who knows :-), but this one's a good one: +1. –  draks ... Jun 9 '12 at 11:59
    
Thanks for the edits @draks ! –  Raymond Manzoni Feb 21 '13 at 21:42

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