2
$\begingroup$

For two integers $a,b$ such that $b \nmid a$, define the "partial zeta function" $$\zeta^{a,b}(s) = \sum_{n \equiv a \, \bmod \, b} n^{-s}, \; \; \mathrm{Re}[s] > 1.$$ The series extends over negative $n$ as well as positive.

I find that these series have simple closed forms at even integers $k$ that are algebraic multiples of $\zeta(k).$ For instance $$\zeta^{1,5}(2) = \frac{10 + 2\sqrt{5}}{125} \pi^2$$ and $$\zeta^{1,5}(4) = \frac{52 + 20\sqrt{5}}{9375} \pi^4.$$ Could anyone explain how to calculate $\zeta^{a,b}$ in terms of $\zeta$?

1 Answers 1

1

According to your notation $$\zeta^{1,5}(2)=\sum_{n\geq 0}\frac{1}{(5n+1)^2}+\sum_{n\geq 0}\frac{1}{(5n+4)^2} = \frac{1}{25}\left[\psi'\left(\frac{1}{5}\right)+\psi'\left(\frac{4}{5}\right)\right]$$ that simplifies to the given expression due to the reflection formula for the trigamma function $$ \psi'(z)+\psi'(1-z) = \frac{\pi^2}{\sin^2(\pi z)}. $$ The second identity can be proved in a similar way, by differentiating twice the previous identity and deriving the reflection formula for the $\psi'''$ function. In general $$ \zeta^{1,p}(2) = g_2\left(\frac{\pi}{p}\right),\qquad g_2(z)=\left(\frac{z}{\sin z}\right)^2,$$ $$ \zeta^{1,p}(4) = g_4\left(\frac{\pi}{p}\right),\quad g_4(z)= \frac{z^4(2+\cos(2z))}{3\sin^4(z)}.$$