I would like to ask a question which has kept me a bit nervous for some time.

As is well-known, \[ - \frac{\zeta'}{\zeta}(2) = \sum_{p} \frac{\Lambda(p)}{p^{2}}, \] where $\Lambda$ is the Von-Mangoldt function.

On the other hand, we can express the fracion $\frac{\zeta'}{\zeta}$ in terms of the zeros of the zeta-function; that is, \[ - \frac{\zeta'}{\zeta}(s) = - \sum_{\rho}(1/(s - \rho) + 1/\rho) + Q(s) \] where $Q(s)$ is some function which is easy to deal with and nothing to do with zeros.

Therefore, we can write \[ - \sum_{\rho}(1/(2 - \rho) + 1/\rho ) + Q(2) = - \frac{\zeta'}{\zeta}(2) = \sum_{p} \frac{\Lambda(p)}{p^{2}}. \]

The question: does the truth/false of the Riemann hypothesis affect the value $(\zeta'/\zeta)(2)$?

Simply put, assume the Riemann hypothesis is true. Then \[ -\frac{\zeta'}{\zeta}(2) = - \sum_{true, \text{Re}(\rho) = 1/2} (1/(2 - \rho) + 1/\rho ) + Q(2) \]

On the other hand, if the Riemann hypothesis is not true, then we have \[ \begin{split} -\frac{\zeta'}{\zeta}(2) & = - \sum_{false, \text{Re}(\rho) = 1/2}(1/(2 - \rho) + 1/\rho ) - \sum_{\text{Re}(\rho) \not= 1/2}(1/(2 - \rho) + 1/\rho ) + Q(2) \\ & \equiv - \sum_{false, \text{Re}(\rho) = 1/2}(1/(2 - \rho) + 1/\rho ) + E(2) + Q(2) \end{split} \]

Whether RH is true or not, the value $-\frac{\zeta'}{\zeta}(2)$ on the left is unchanged. The question is, are the sums \[ \sum_{true, \text{Re}(\rho) = 1/2} \] and \[ \sum_{false, \text{Re}(\rho) = 1/2} \] really the same? Could the latter be a function of zeros off the line?

\[ \sum_{false, \text{Re}(\rho) = 1/2} =: F(E(2)) \quad ? \]

If they are the same, then RH actually would affect the value of $- (\zeta'/\zeta)(2)$. That can't be.

"They are not the same" should be the correct answer, I guess.

Any idea?


  • $\begingroup$ First off, $-\frac{\zeta'}{\zeta}(2) = \sum_{n = 1}^{\infty} \frac{\Lambda(n)}{n^2} = \sum_p \sum_{k = 1}^{\infty} \frac{\log p}{p^{2k}} = \sum_p \frac{\log p}{p^2 - 1}$. Secondly, this is a constant, so it makes no sense to ask whether the validity of the Riemann hypothesis changes its value. It's a fixed number! $\endgroup$ – Peter Humphries Oct 21 '17 at 11:28
  • 1
    $\begingroup$ Exactly, it is like asking if RH affects the value of $\pi$. It doesn't, clearly. It makes much more sense to ask how the set of values $\frac{\zeta'}{\zeta}(2),\frac{\zeta'}{\zeta}(4),\frac{\zeta'}{\zeta}(6),\ldots$ (computed with some accuracy) affects the distribution of the zeroes of $\zeta$. $\endgroup$ – Jack D'Aurizio Oct 21 '17 at 14:00
  • $\begingroup$ A simple criterion for the RH outside the critical strip is given by the Li criterion. Also see how the sequence of derivatives $F^{(k)}(s_0)$ for some $\Re(s_0) > 1$ where $F(s) = \frac{1}{\zeta(s)}$ or $\frac{\zeta'(s)}{\zeta(s)}+\frac{1}{s-1}$ encodes the abscissa of convergence of $\sum_{n=1}^\infty \mu(n) n^{-s}$ and $\sum_{n=1}^\infty (\Lambda(n)-1)n^{-s}$ and hence the RH. $\endgroup$ – reuns Oct 21 '17 at 19:36
  • $\begingroup$ @PeterHumphries If what you say is correct, then the error in PNT $\pi(x) - Li(x)$ is irrelevant of the RH? $\pi(x)$ of course depends on prime distribution, and if $\pi(x)$ takes the same values for all $x$ regardless of the RH, then we get the same error terms in PNT (Li(x) definitely is irrelevant of the RH)! $\endgroup$ – Grown pains Oct 21 '17 at 21:54
  • $\begingroup$ $\pi(x) - \mathrm{Li}(x)$ for fixed $x$ is a constant, and its size tells you nothing about RH. As $x$ tends to infinity, on the other hand, the validity of RH of course affects the behaviour of this. $\endgroup$ – Peter Humphries Oct 21 '17 at 21:57

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