6
$\begingroup$

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.

$\endgroup$
0

4 Answers 4

7
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ Thank you. I made a computational mistake and now agree with your answer. $\endgroup$
    – joro
    Mar 16, 2012 at 9:15
  • $\begingroup$ 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? $\endgroup$
    – joro
    Mar 16, 2012 at 15:52
8
$\begingroup$

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).

$\endgroup$
5
  • $\begingroup$ 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 $1$ $\endgroup$
    – juan
    Mar 16, 2012 at 8:44
  • $\begingroup$ Thank you. I made a computational mistake and now agree with Micah's answer. $\endgroup$
    – joro
    Mar 16, 2012 at 9:14
  • $\begingroup$ 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? $\endgroup$
    – joro
    Mar 16, 2012 at 12:50
  • $\begingroup$ 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 $L(s,\overline{\chi})$. $\endgroup$
    – juan
    Mar 16, 2012 at 17:54
  • $\begingroup$ Thanks for the books, juan. Dumb me, now i see why the 2 sums are equal at the possible zeros whatever they might be. $\endgroup$
    – joro
    Mar 25, 2012 at 11:22
7
$\begingroup$

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.

$\endgroup$
0
5
$\begingroup$

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

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

$\endgroup$
1
  • $\begingroup$ 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 $\sum_{\rho} \frac{1}{\rho (1{-}\rho)}=\sum_{\rho} \frac{1}{|\rho|^2} = 2 + \gamma - \log 4\pi$. $\endgroup$
    – joro
    Mar 15, 2012 at 14:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.