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Let $ G(s) := \frac{\zeta'(s)}{\zeta(s)} + \frac{\zeta'(1-s)}{\zeta(1-s)}$ where $s$ is not a zero of zeta.

$G$ has real zeros and a pair of complex zeros near $\frac12 \pm 6i$.

Partial results:

By differentiating $\log{\xi(s)} - \log{\xi(1-s)}$ one gets:

$$ G(s) = -1/2\psi(1/2+s/2)+1/2\pi\tan(\pi s/2) -1/2 \psi(s/2) +\log\pi \qquad (1) $$

when $s$ is not a zeta zeros. Probably (1) can be simplified since $\Re G(s) = \Re\left(-1/(s-1)+1/s+\log\pi -(\psi(s/2)+\psi((1-s)/2+1)/2\right)$

Are there other complex zeros of (1), especially in the critical strip?

X-Ray of $G(s)$ using (1):

Improve this question
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  • $\begingroup$ $\xi(s) = \xi(1-s)$, isn't it? What is your definition of $\xi$? $\endgroup$
    – Marc Palm
    Jun 19, 2014 at 12:59
  • $\begingroup$ @plusepsilon.de IIRC it is Riemann xi: en.wikipedia.org/wiki/Riemann_Xi_function $\endgroup$
    – joro
    Jun 19, 2014 at 13:02
  • $\begingroup$ @plusepsilon.de Yes, your equality holds. So by differentiating the logs the expression must be $0$, which how I derived the G(s). $\endgroup$
    – joro
    Jun 19, 2014 at 13:06
  • $\begingroup$ @plusepsilon.de I meant: first express xi in terms of zeta and then... $\endgroup$
    – joro
    Jun 19, 2014 at 13:16
  • $\begingroup$ How do cancel the Euler products? Seems like you only got the $\Gamma$ factor. $\endgroup$
    – Marc Palm
    Jun 19, 2014 at 16:30

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