# Is this sum of reciprocals of zeta zeros correct?

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.

-

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

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

-
Thank you. I made a computational mistake and now agree with your answer. –  joro Mar 16 '12 at 9:15
Per juan's answer your sum is unconditionally $=\sum_{\rho} \rho^{-1}+\overline{\rho}^{-1}$ (the constant depends if one takes only Im(s)>0 or not). Since on the critical line the terms are equal does this mean the difference of the sums over potential Siegel zeros must vanish unconditionally? –  joro Mar 16 '12 at 15:52

The series $\sum_\rho \rho^{-1}$ over the non-trivial zeros is not absolutely convergent, this is proved in Davenport p. 80. But as Davenport says and proves in page 81-82 the series converges conditionally provided one groups together the terms from $\rho$ and its conjugate $\overline{\rho}$. And the value of the sum can be given, independently of RH, as the constant $-B$ where $B = -\frac12 \gamma-1+\frac12\log4\pi$. (This value was known to Riemann, as Siegel says in his paper about the Riemann Nachlass).

-
In Lagarias formula the sum is over the non trivial zeros repeated according to its multiplicity. In fact the formula is the Mittag-Leffler expansion of the meromorphic function $\frac{\hat\zeta'(s)}{\hat\zeta(s)}$ that has poles just at the non-trivial zeros and at $0$ and $1$ –  juan Mar 16 '12 at 8:44
Thank you. I made a computational mistake and now agree with Micah's answer. –  joro Mar 16 '12 at 9:14
Looks like Micah's sum $\sum_{\rho} \rho^{-1}+(1-\rho)^{-1}$ is exactly $\sum_{\rho} \rho^{-1}+\overline{\rho}^{-1}$ (Davenport's $B$ sums only Im(s)>0 and this explains the factor of $2$. Since on the critical line the terms are equal does this mean the difference of the sums over potential Siegel zeros must vanish unconditionally? –  joro Mar 16 '12 at 12:50
In fact the two sums are equal $\sum_\rho(\rho^{-1}+(1-\rho)^{-1})= \sum_\rho (\rho^{-1}+\overline{\rho}^{-1})$. But I (or Davenport) am speaking about $\zeta(s)$. The case of $L(\chi,s)$ is a little more complicate. In this case the zeros are symmetric with respect to the line $\sigma=\frac12$ but not respect the real axis. The determination of the constant $B(\chi)$ is recent, due to Vorhauer in 2006. Davenport do not include it. You must see the book by Montgomery and Vaughan, Chapter 10. In the sum you must consider the zeros of $L(s,\chi)$ and those of $L(s,\overline{\chi})$. –  juan Mar 16 '12 at 17:54
Many thanks juan. –  joro Mar 17 '12 at 7:37

Edit: I endorse Juan's answer to the original question. The sum $\displaystyle{\sum_{\rho} \tfrac{1}{|\rho|}}$, running over the non-trivial zeros $\rho$ of $\zeta(s)$, is known to diverge, so at best $\displaystyle{\sum_{\rho} \tfrac{1}{\rho}}$ is conditionally convergent so you cannot re-arrange the terms.

In your second to last displayed equation, you removed the assumption that the sum runs over pairs of zeros $\rho$ and $1-\rho$. So it seems that Lagarias' result can be used to evaluate the sum $$\sum_{\rho} \frac{1}{\rho (1{-}\rho)}.$$

As you observed, assuming the Riemann Hypothesis $1-\rho =\overline{\rho}$ for any non-trivial zero $\rho$ of $\zeta(s)$. This implies that

$$\sum_{\rho} \frac{1}{\rho (1{-}\rho)}=\sum_{\rho} \frac{1}{|\rho|^2}.$$

-
Thank you. Computing $\sum_{\rho} \frac{1}{|\rho|^2}$ is equivalent to RH according to aimath.org/pl/rhequivalences 5.3. I might be missing something but if RH holds $\sum_{\rho} \frac{1}{\rho (1{-}\rho)}=\sum_{\rho} \frac{1}{|\rho|^2} = 2 + \gamma - \log 4\pi$. –  joro Mar 15 '12 at 14:19

These identities are not mysterious. They are simply the fact that the Riemann Zeta function has a Weierstrass product like any other meromorphic function of finite exponential order. Note here that $f'/f$ is called logarithmic derivative for a reason;)

Then it follows immediately

1) Yes, the zeros of the completed Riemann zeta function are exactly the nontrivial ones.

2) If RH hold, they come in pairs $\rho = 1/2 \pm \mathrm{i}t$ for $t>0$.

A suggestion for computing the sum:

Let $\Omega_T$ be the boundary of $-T \leq Im s \leq T$ and $-1/2 < Re s < 3/2$. For an approximation consider the integral $$\frac{1}{2 \pi i} \int\limits_{\Omega_T} s^{-1} \frac{\zeta'(s)}{\zeta(s)} d \; s = \sum\limits_{-T \leq Im \rho \leq T} \frac{1}{\rho} +$$ some contribution coming from the poles, which are slightly delicate for $s=0$, since you encounter a double pole. I am pretty sure that you have missed that, and that this is why your computation fails.

Wikipedia tells you that $$\zeta(s) = \frac{2^{s-1}}{s-1}-2^s \int_0^{\infty}\frac{\sin(s\arctan t)}{(1+t^2)^\frac{s}{2}(\mathrm{e}^{\pi\,t}+1)}\,\mathrm{d}t,$$ is pretty convenient for computing $\zeta$ numerically.

-
Thank you...... –  joro Mar 16 '12 at 12:14