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In a (side) comment to this question it is claimed that for a certain domain the following holds:

$$-\frac{\zeta'(s)}{\zeta(s)}= \sum_{n=2}^\infty \frac{\psi(n+1/2)-\psi(n-1/2)} {n^{s}} = \sum_{n=2}^\infty \left(1-\sum_\rho \frac{(n+1/2)^\rho-(n-1/2)^\rho}{\rho\, n^{s}}\right)$$

The $\rho$'s are both the trivial and (pairs of) the non-trivial zeros of $\zeta(s)$.

When I isolate the series for the trivial zeros (at $-2k$) from the RHS, I get to:

$$\sum_{k=2}^{\infty}\frac{\frac12\ln\left(2k+3\right)-\frac12\ln\left(2k-3\right)+\frac32\ln\left(2k-1\right)-\frac32\ln\left(2k+1\right)}{k^s}$$

which is equal to: $$f(s)=\sum_{k=2}^{\infty}\frac{\coth^{-1}\left(\frac23\,k\right)-3\coth^{-1}\left(2\,k\right)}{k^s}$$

and seems to converge for all $\Re(s)>-1$.

Could there exist a closed form for this series (e.g. in terms of $\zeta'(s)\,$)?

P.S.:

Numerical evidence suggests that all complex zeros of $f(s) \pm f(1-s)$ reside on the line $\Re(s)=\frac12$ (for the domain $-1 < \Re(s) < 2$), however the distribution of the imaginary parts on the line itself seems quite irregular.

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    $\begingroup$ Here $a = 1/2$, but you can take any $a \in (0,1)$, since $\psi(n+a)-\psi(n-a)$ doesn't depend on $a \in (0,1)$. But the trivial zeros part of the series does depend on $a$. Expanding $n^{-s}(n+a)^\rho= \sum_{m \ge 0} {\rho \choose m}a^m n^{-s+\rho-m}$ you'll obtain something like $\sum_{m \ge 0, k \ge 1} {-2k \choose m} \frac{a^{-2m}-(-a)^{-2m}}{2k}\zeta(s+m+2k)$. Maybe you can express it as a series in $\frac{\Gamma'((s+l)/2)}{\Gamma((s+l)/2)}$ $\endgroup$ – reuns Dec 5 '17 at 13:01
  • $\begingroup$ Thanks @reuns. Have checked other values of $a$, but unfortunately they do not yield any closed form. Also tried $f(s) + f(1-s)$, since $\frac{\zeta'(s)}{\zeta(s)}+\frac{\zeta'(1-s)}{\zeta(1-s)}$ does have a nice closed form expression. Also without success. $\endgroup$ – Agno Dec 5 '17 at 19:14

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