# Is it possible to simplify $\psi^{(2)}(\frac18)$ or $\psi^{(2)}(\frac pq)$?

Is it possible to simplify $$\psi^{(2)}(\frac18)$$, where $$\psi$$ denotes the polygamma function?
Or more generalized, $$\psi^{(2)}(\frac pq)$$ and $$\psi^{(2n)}(\frac pq)$$?

Background
Noticing there is a formula for $$\psi(\frac pq)$$, where $$p,q\in\mathbb{N}$$ and the following formulae:
$$\psi^{(2)}(\frac12)=-14\zeta(3)$$, $$\psi^{(2)}(\frac13)=-26\zeta (3)-\frac{4\pi^3}{3\sqrt{3}}$$, $$\psi^{(2)}(\frac14)=-2\pi^3-56\zeta(3)$$ and $$\psi^{(2)}(\frac16)=-182\zeta (3)-4\sqrt{3}\pi^3$$, it is natural to ask if there are general formulae.
Attempt
$$\psi^{(2)}\left(\frac pq\right)=\int_0^1\frac{\ln^2 t}{t-1}t^{p/q-1}dt\\=q^3\int_0^1\frac{\ln^2 u}{u^q-1}u^{p-1}du$$ I tried to use contour integration to solve this integral but failed.

• Removing the syntactic sugar, the OP is asking for a closed form for $$\sum_{n\geq 0}\frac{1}{(qn+p)^3}.$$ – Jack D'Aurizio Oct 12 '18 at 19:51

This is an answer showing that why $$\psi^{(2n)}\left(\frac pq\right)$$ has a closed-form when $$q=2,3,4,6$$ and conjecturing it does not have a general closed-form when $$p=5$$ or $$p\ge7$$. (Assuming $$L(s,\chi)$$ is not a closed-form)
1.$$L(2n+1,\chi)$$ has a closed-form when $$\chi(-1)=-1$$
(Where $$L$$ denotes the $$L$$-series)
First, denote $$f(x,\chi)$$ the analytic continuation of $$\displaystyle\sum_{n=0}^\infty \chi(n)x^n$$. In this question, I showed that $$f(x)=-\chi(-1)f\left(\frac1x\right)$$.
Noticing $$\int_0^1 x^{n-1}(-\ln x)^{s-1}\frac{dx}{\Gamma(s)}=n^{-s},$$ one can prove that it is valid to change the position of $$\sum$$ and $$\int$$: $$L(2s+1,\chi)=\int_0^1\sum_{n=0}^\infty\chi(n)x^{n-1}(-\ln x)^{2s}\frac{dx}{\Gamma(s)}\\ =\frac{1}{2\Gamma(s)}\int_0^\infty\frac{f(x)}{x}(-\ln x)^{2s}dx$$ Since $$f(x)$$ can be expressed in form $$\frac{P(x)}{Q(x)}$$, where $$P$$ and $$Q$$ are polynomial funtions, the integral has a closed-form according to Residue Theorem.
2.$$L(n,\chi)$$ has a closed-form when $$\chi$$ is a principal character
If $$n$$ is a prime, it is trivial.
If $$n=p_1^{\alpha_1}\cdots p_{\tau}^{\alpha_\tau}$$, it is always possible to add and subtract some terms to make the series $$\zeta(n)$$. Furthermore, $$L(n,\chi)=\frac pq\zeta(n)$$
3.It is possible to deduce $$\psi^{(2n)}\left(\frac pq\right)$$ from the values of $$L$$-series when $$q=2,3,4,6$$.
For simpleness, I will discuss $$\zeta\left(n,\frac pq\right)$$ instead of $$\psi$$ here.
$$2^nL(n,\chi_{2,1})=\zeta(n,\frac12)$$,
$$3^nL(n,\chi_{3,1})=\zeta(n,\frac13)+\zeta(n,\frac23)$$, $$3^nL(n,\chi_{3,2})=\zeta(n,\frac13)-\zeta(n,\frac23)$$
$$4^nL(n,\chi_{4,1})=\zeta(n,\frac14)+\zeta(n,\frac34)$$, $$4^nL(n,\chi_{4,2})=\zeta(n,\frac14)-\zeta(n,\frac34)$$
$$6^nL(n,\chi_{6,1})=\zeta(n,\frac16)+\zeta(n,\frac56)$$, $$6^nL(n,\chi_{6,2})=\zeta(n,\frac16)-\zeta(n,\frac56)$$
Solving this simultaneous equation gives the values of $$\zeta\left(n,\frac pq\right)$$. (The values of $$L$$-series in LHS is known when $$n$$ is odd)
4.Conjecturing it does not have a general closed-form when $$p=5$$ or $$p\ge7$$
This post shows that $$\varphi(n)>2$$ when $$n>6$$. It is also obvious that $$\varphi(5)>2$$.
There is $$\frac12\varphi(k)+1$$ values known in $$L(2s+1,\chi)$$, and we have $$\varphi(k)$$ variables. Since $$\varphi(k)>2$$ for $$k=5$$ and $$k>6$$, $$\varphi(k)>\frac12\varphi(k)+1$$. It is not possible to deduce $$\zeta(2s+1,\frac pq)$$ via $$L$$-series.