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In my mathematics course in Uni. (I'm a physics student) my prof. gave us the following exercise: to express the Hurwitz Zeta function $\zeta(2k+1,\frac{1}{4})$ with $k=1,2,3,\dots$ in terms of the Riemann zeta function. He says there is a closed form for this, something like $\zeta(2k+1,\frac{1}{4})=C(k)\zeta(2k+1)$. With $C(k)$ some elementary function of $k$.

After some basic calculations I found the following

$\zeta(2k+1,\frac{1}{4})=2^{2k+1}(2^{2k+1}-1)\zeta(2k+1)-\zeta(2k+1,\frac{3}{4})$

but I don't know what to do about $\zeta(2k+1,\frac{3}{4})$. Trying the same method on this function, I stumble upon $\zeta(2k+1,\frac{1}{8})$. And I'm going in a loop getting nothing like my prof. said. Also I've looked through books and articles on the Hurwitz function and found nothing.

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2 Answers 2

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It can't be written as a scaled up version of the zeta function by elementary functions, however $$\zeta(2k+1,\frac{1}{4})=2^{2k}(2^{2k+1}-1)\zeta(2k+1)+\frac{(-4)^kE_{2k}\pi^{2k+1}}{2(2k)!}$$

Where $E_k$ are euler numbers, with the first few

$E_0=1$

$E_2=-1$

$E_4=5$

$E_6=-61$

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\begin{equation} \begin{array}{c} \left. \begin{array}{c} \zeta (2n+1,\frac{1}{4}) \\ \zeta (2n+1,\frac{3}{4})% \end{array}% \right\} =2^{2n}(2^{2n+1}-1){\zeta }(2n+1) \\ \pm \frac{1}{2\pi }\left( 2n+2+4^{2n+2}\right) {\zeta }(2n+2)-2\sum \limits_{l=0}^{n-1}4^{2n-2l}{{{\zeta }(2n-2l)\zeta }}(2l+2)% \end{array} \tag*{(1)} \end{equation}

\begin{equation} \begin{array}{c} \left. \begin{array}{c} \zeta (2n+1,\frac{1}{3}) \\ \zeta (2n+1,\frac{2}{3})% \end{array}% \right\} =\frac{{3^{2n+1}-1}}{2}{\zeta (2n+1)} \\ \pm \frac{\sqrt{3}}{2\pi }\left( \left( 2n+2+3^{2n+2}\right) {\zeta }% (2n+2)-2\sum\limits_{l=0}^{n-1}3^{2n-2l}{{{\zeta }(2n-2l)\zeta }}% (2l+2)\right)% \end{array} \tag*{(2)} \end{equation}

\begin{equation} \begin{array}{c} \left. \begin{array}{c} \zeta (2n+1,\frac{1}{6}) \\ \zeta (2n+1,\frac{5}{6})% \end{array}% \right\} =\frac{{6^{2n+1}-{3^{2n+1}}-{{2^{2n+1}}+1}}}{2}{\zeta (2n+1)} \\ \pm \frac{1}{2\sqrt{3}\pi }\left( 6^{2n+2}-3^{2n+2}\right) {\zeta }% (2n+2)-2\sum\limits_{l=0}^{n-1}\left( 6^{2n-2l}-3^{2n-2l}\right) {{{\zeta }% (2n-2l)\zeta }}(2l+2)% \end{array} \tag*{(3)} \end{equation}

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  • $\begingroup$ Can you give a source of your formulas, please? $\endgroup$
    – Ben Hur
    Aug 22, 2023 at 13:41

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