- $(\zeta'/ \zeta)(2)$ and zeros of the zeta-function

187 Views Asked by At

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.