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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?

Thanks.

Grown pains
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  • 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! – Peter Humphries Oct 21 '17 at 11:28
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    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 $\zeta$. – Jack D'Aurizio Oct 21 '17 at 14:00
  • 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. – reuns Oct 21 '17 at 19:36
  • @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)! – Grown pains Oct 21 '17 at 21:54
  • $\pi(x) - \mathrm{Li}(x)$ 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. – Peter Humphries Oct 21 '17 at 21:57
  • @PeterHumphries >As x tends to infinity, ... Maybe you missed my point. If $\pi(x)$ takes the same values for all $x$ regardless of RH, then the behavior of $\pi(x) - Li(x)$ is the same regardless of RH. Besides, I have seen $Lambda(n)$ expressed in terms of zeros (by Landau, I guess). Thus, $- (\zeta' / \zeta)(2)$ should be dependent of zeros? – Grown pains Oct 22 '17 at 07:53
  • I don't think I understand what you're trying to say. The conjectured size of $\pi(x)$, as a function of $x$, is equivalent to RH. The size of $\pi(x)$ for fixed $x$ is a constant. What do you mean by "takes the same values for all $x$ regardless of RH"? – Peter Humphries Oct 22 '17 at 15:17
  • While $-\frac{\zeta'}{\zeta}(2)$ can be expressed in terms of zeroes, it can also be expressed as $\frac{6}{\pi^2} \sum_{n = 2}^{\infty} \frac{\log n}{n^2}$. This is a constant! – Peter Humphries Oct 22 '17 at 15:17
  • Here's an example: the sum $\sum_{\rho} \left(\frac{1}{1 - \rho} + \frac{1}{\rho}\right) = 2\sum_{\rho} \Re\left(\frac{1}{\rho}\right)$ is a constant defined in terms of the zeroes of $\zeta(s)$. But this constant is equal to $\gamma_0 + 2 - \log 4\pi$, where $\gamma_0$ is the Euler-Mascheroni constant. – Peter Humphries Oct 22 '17 at 15:23
  • Here is what I meant informally put. The prime sequences should look different depending on the validity of RH. If they (that is, prime sequence under RH and prime sequence under RH false) do look alike exactly, then pi(x) - Li(x) got to be the same too for all x. But now, I think I see your point. RH wrong -> $\gamma_{0} + 2 - \log 4 \pi$ changes value can't be. Another possibility I have been thinking is that depending on RH, the series over zeros on the critical line may change value, so that it fixes discrepancy caused by zeros off the line? – Grown pains Oct 24 '17 at 13:10
  • There is only one truth. I mean, it's totally nonsense comparing a relation under RH with another under RH false. A friend of mine has shown me an argument varying zeros off the line in some relation and see what he can say about conditions which those zeros off the line must satisfy. As you can see, however, it's really hard for me to think properly and correctly by myself. – Grown pains Oct 24 '17 at 13:26
  • I edited my question to make clear the point. Is the $\sum_{true, re \rho = 1/2}$ (the sum over zeros on the critical line under RH being true) equal to $\sum_{false, re \rho = 1/2}$ (that under RH being false) ? Your opinion, I would like to hear. I somewhat have begun to think that I can give explanation for each case whether they are the same or not, which again really confuses me. – Grown pains Nov 12 '17 at 22:04

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