Numerical evidence suggests that all complex zeros (real ones exist as well) of:

$$\frac{\zeta'}{\zeta}(s) \pm \frac{\zeta'}{\zeta}(1-s)$$

reside on the critical line with $\Re(s)=\frac12$.

I made some progress by taking:

(1) $\zeta(s):=\chi(1-s)\,\zeta(1-s), \, \, \chi(s)= \Gamma \left( s \right) \cos \left(\frac12\pi s \right) 2 \left( 2\pi \right) ^{-s}$

(2) $\zeta(s):= -\dfrac{\zeta'(1-s)+\chi(s)\,\zeta'(s)}{\chi'(s)}$ (derived from Apostol's paper found here).

and then the formulae can be rewritten into:

\begin{align*} \frac{\zeta'}{\zeta}(s) + \frac{\zeta'}{\zeta}(1-s) &= -\dfrac{\chi'}{\chi}(s) \\ \\ \frac{\zeta'}{\zeta}(s) - \frac{\zeta'}{\zeta}(1-s) &= -\dfrac{\chi'}{\chi}(s) \cdot \dfrac{\chi(s)\,\zeta'(s)-\zeta'(1-s)}{\chi(s)\,\zeta'(s)+\zeta'(1-s)} \\ \end{align*}

The (+) version only has a single pair of zeros at $\frac12 \pm 6.2898359888369027796...$ and shows a monotonically increasing absolute value from that point onwards (note that the expected poles at the non-trivial zeros $\rho$ are all annihilated by $\zeta'(1-s)+\chi(s)\,\zeta'(s)$) that also induces the $\rho$s).

The (-) version does have a pole at each $\rho$, however these appear to be always separated by a single new zero that apparently always resides on the critical line. Could the latter be proved?


  • $\begingroup$ Remotely related: mathoverflow.net/questions/134029/… $\endgroup$
    – joro
    Commented May 10, 2015 at 18:47
  • $\begingroup$ Thanks Joro. This one mathoverflow.net/questions/93323/are-all-zeros-of-ζks±ζk1−s-on-the-critical-line-k-k-th-derivative is also related, although in that question there are zeros outside the strip (contrary to the question above). $\endgroup$
    – Agno
    Commented May 10, 2015 at 19:50
  • $\begingroup$ What do you about the expected poles of "+"? Zero off the line will induce pole in the LHS and in the RHS. Or am I missing something? $\endgroup$
    – joro
    Commented May 11, 2015 at 6:50
  • $\begingroup$ I am not sure the RHS have poles at zeros. If this is true, this will give alternative computation of zeros -- look for poles of RHS, which is much simpler than zeta. $\endgroup$
    – joro
    Commented May 11, 2015 at 11:52
  • $\begingroup$ Formula not involving zeta having zeros/poles at zeta zeros might be quite interesting. Agno, do you get poles of the "+" RHS at zeta zeros? $\endgroup$
    – joro
    Commented May 11, 2015 at 12:21

1 Answer 1


See the following papers which treat this and similar questions.

  • MR1986257 (2004c:11152) Reviewed Saidak, Filip(3-CALG-MS); Zvengrowski, Peter(3-CALG-MS) On the modulus of the Riemann zeta function in the critical strip. (English summary) Math. Slovaca 53 (2003), no. 2, 145–172.

  • MR3277049 Reviewed Matiyasevich, Yu.(RS-AOS2); Saidak, F.(1-NCG); Zvengrowski, P.(3-CALG-MS) Horizontal monotonicity of the modulus of the zeta function, L-functions, and related functions. Acta Arith. 166 (2014), no. 2, 189–200.

  • $\begingroup$ Many thanks for these references, Peter. Very helpful. I have reviewed both documents now and immediately recognised the threshold at $|t|=6.2898..i$ (they put this at $2\,\pi+1$) below which the monotonicity breaks down (but of course that lower region can be inspected on the location of zeros by computer). The proof of the horizontal monotonicity for $\sigma < 0$ is nice, however it now also becomes clear to me that an extended proof towards the region $0 \le \sigma \le \frac12$ actually implies the RH. Could therefore a proof of my conjecture be constructed assuming the RH a priori? $\endgroup$
    – Agno
    Commented May 10, 2015 at 20:04

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