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While attempting to solve the question Alternating harmonic sum $\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k$ we have discovered a following puzzling identity: \begin{equation} Li_q(-1) = \left(2^{1-q} - 1\right) \zeta(q)=\frac{(-1)^{q-1}}{(q-1)! 2^q} \left( \Psi^{(q-1)}(\frac{1}{2}) - \Psi^{(q-1)}(1)\right) \end{equation} for $q=2,3,4,\cdots$. As far as I know no closed form expressions are known for zeta function values at odd integers. The relationship above does give some closed form expression. Now, the question would be is this relationship something trivial or are there any generalizations of that?

Note: Using the series representation of the polygamma function the above relation simply reduces to: \begin{equation} \Psi^{(q-1)} \left(\frac{1}{2}\right) = \left(2^q-1\right) \Psi^{(q-1)}\left(1\right) =\left(2^q-1\right) (-1)^q (q-1)! \zeta(q) \end{equation} Is this a trivial identity or not ?

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  • $\begingroup$ Depends on the interpretation of trivial, e.q. simply insert dlmf.nist.gov/5.15.E2 into dlmf.nist.gov/5.15.E3 $\endgroup$
    – gammatester
    Aug 31, 2016 at 12:06
  • $\begingroup$ Concerning your first question: Why would you consider $\Psi^{(q-1)}(\frac{1}{2})$ a closed form but not $\zeta(q)?$ $\endgroup$
    – gammatester
    Aug 31, 2016 at 13:19
  • $\begingroup$ @gammatester: The first question was asked very hastily without thinking through. I take it back. The above simply relates all derivatives of the Gamma function at one half to to values of zeta function at integers. If we had an independent way to compute the former that certainly this would pave the way to finding zeta function values at odd integers. By the way do you know if there is any progress in solving this very problem? $\endgroup$
    – Przemo
    Aug 31, 2016 at 13:27
  • $\begingroup$ Unfortunately I do not know about any progress. $\endgroup$
    – gammatester
    Aug 31, 2016 at 14:03

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