Is there a nice identity known for $\frac{\zeta(k- \tfrac{1}{2}) \zeta(2k -1)}{\zeta(4k -2)}?$ (I'm dealing with half-integral $k$.) Equally, an identity for $\frac{\zeta(s) \zeta(2s)}{\zeta(4s)}$ would do ;)
Identity for $\zeta(k- 1/2) \zeta(2k -1) / \zeta(4k -2)$?
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1If there were a simplification not involving odd $s$ one could evaluate for example zeta(3), which is not now known in closed form. – 2012-11-28
1 Answers
Let $F(s) = \frac{\zeta(s)\zeta(2s)}{\zeta(4s)}.$ Then clearly the Euler product of $F(s)$ is $F(s) = \prod_p \frac{\frac{1}{1-1/p^s}\frac{1}{1-1/p^{2s}}}{\frac{1}{1-1/p^{4s}}}= \prod_p \left( 1 + \frac{1}{p^s} + \frac{2}{p^{2s}} + \frac{2}{p^{3s}} + \frac{2}{p^{4s}} + \frac{2}{p^{5s}} + \cdots\right).$ Now introduce $ f(n) = \prod_{p^2|n} 2.$ It follows that $ F(s) = \sum_{n\ge 1} \frac{f(n)}{n^s}.$ We can use this e.g. to study the average order of $f(n)$, given by $ \frac{1}{n} \sum_{k=1}^n f(n).$ The function $F(s)$ has a simple pole at $s=1$ and the Wiener-Ikehara-Theorem applies. The residue is $\operatorname{Res}_{s=1} F(s) = \frac{15}{\pi^2}$ so that finally $ \frac{1}{n} \sum_{k=1}^n f(n) \sim \frac{15}{\pi^2}.$ In fact I would conjecture that we can do better and we ought to have $ \frac{1}{n} \sum_{k=1}^n f(n) \sim \frac{15}{\pi^2} + \frac{6}{\pi^2}\zeta\left(\frac{1}{2}\right) n^{-1/2}.$
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0nice thoughts, thank you! – 2012-12-02