Is there a general summation formula for the polygamma function at $z=1/2$? i.e $\psi^{(s)}(\frac{1}{2})$ for all s.

Question

For $s>0$ one has $ \psi^{(s)}(\frac{1}{2}) = s! \cdot \zeta(s+1, \frac{1}{2}) \cdot (-1)^{s+1} $.

E.g.
$ \psi^{(1)}(\frac{1}{2}) = 3 \cdot \zeta(2) $
$ \psi^{(2)}(\frac{1}{2}) = -14 \cdot \zeta(3) $
$ \psi^{(3)}(\frac{1}{2}) = 90 \cdot \zeta(4) $
etc.

However, Mathematica gives
$ \psi^{(-1)}(\frac{1}{2}) = \frac{\ln(\pi)}{2} $
$ \psi^{(-2)}(\frac{1}{2}) = \ln(A^\frac{3}{2}2^\frac{5}{24}\pi^\frac{1}{4}), $ where $A$ is the Glaisher-Kinkelin Constant
$ \psi^{(-3)}(\frac{1}{2}) = \ln(A^\frac{1}{2}2^\frac{1}{16}\pi^\frac{1}{16}) + \frac{7 \zeta(3)}{32\pi^2} $

and I'm not sure how these are obtained since negative integers plugged into the formula above yields wonky results.

Answer 1

First off we have for reference this formula which states for all $n\in\Bbb N$ $$ \psi ^{(n)}\left(\tfrac{1}{2}\right)=(-1)^{n+1} \left(2^{n+1}-1\right) n! \zeta (n+1). $$ Note that this is not the same formula as given in your question statement. As for negative values, I was not able to find a closed-form for all $-n\in\Bbb N$ at $z=1/2$ but there is one for $-2n$ with $n\in\Bbb N$ and $z=1/2$ given here. Furthermore, we do have a formula for general $-n$ and $z$ given here which can be evaluated at $z=1/2$. With a little extra work you could further simplify this to the special cases you obtained using Mathematica using the well-known properties of the polygamma function.