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In relation with Evaluating series of zeta values like $\sum_{k=1}^{\infty} \frac{\zeta(2k)}{k16^{k}}=\ln(\pi)-\frac{3}{2}\ln(2) $

From the well-known formula,

For $s$, such that $\Re(s)>1$,

$\displaystyle \zeta(s)=\dfrac{1}{\Gamma(s)}\int_0^{+\infty} \dfrac{x^{s-1}}{e^x-1}dx$

it follows that,

$\displaystyle \sum_{k=1}^{\infty} \frac{\zeta(2k)}{k16^{k}}=4\int_0^{+\infty} \dfrac{\left(\sinh\left(\tfrac{x}{8}\right)\right)^2}{x(e^x-1)}dx$

Is it possible to evaluate directly the latter integral?

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    $\begingroup$ Looks like this with $a= i/8.$ $\endgroup$ Jul 18, 2016 at 13:39
  • $\begingroup$ try abel plana! $\endgroup$
    – tired
    Jul 18, 2016 at 14:01
  • $\begingroup$ @FDP Please, can you explain what's wrong with the answers below? $\endgroup$ Feb 1, 2018 at 18:58

3 Answers 3

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An approach.

Hint. One may set $$ f(s):=4\int_0^\infty \frac{\left(\sinh\left(sx\right)\right)^2}{x(e^x-1)}dx, \quad 0<s<\frac12. \tag1 $$ By differentiating with respect to $s$ one gets $$ \begin{align} f'(s)&=4\int_0^\infty \frac{\sinh\left(2sx\right)}{e^x-1}dx \\\\&=2\int_0^\infty\left(e^{2sx}-e^{-2sx}\right)\sum_{n=1}^\infty e^{-nx}dx \\\\&=2\sum_{n=1}^\infty\int_0^\infty\left(e^{2sx}-e^{-2sx}\right) e^{-nx}dx \\\\&=\sum_{n=1}^\infty\frac{8s}{n^2-4s^2} \\\\&=\frac1s-2\pi\cot(2\pi s) \tag2 \end{align} $$ Integrating $(2)$, with the fact that, as $s \to 0$, $f(s) \to 0$, we get

$$ \int_0^\infty \frac{\left(\sinh\left(sx\right)\right)^2}{x(e^x-1)}dx=\log \left(\frac{2\pi s}{\sin(2\pi s)}\right), \quad 0<s<\frac12, \tag4 $$

from which you deduce the value of the given integral by putting $s:=\color{blue}{\frac18}$, which is $\color{blue}{\ln(\pi)-\frac32\ln 2}$.

Remark. With the general parameter $s$, we have obtained a little bit more than the initial integral.

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    $\begingroup$ Whoever downvoted, please indicate the reason. I upvoted, because I don't see anything wrong with this answer $\endgroup$
    – Yuriy S
    Jul 18, 2016 at 14:15
  • $\begingroup$ @You're In My Eye and Jack D'Aurizio good Job (+1) for all $\endgroup$ Jul 18, 2016 at 14:25
  • $\begingroup$ So he differentiates the integrand. Gets an integral that evaluates to a function that has pole at the value he wants to integrate to and then goes and just permutes the derivative with the integral. This needs some analysis to justify the permutation of the derivative with the integral. It's far from obvious. $\endgroup$
    – zzchan
    Jul 18, 2016 at 14:30
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    $\begingroup$ @zzchan Do you even know what does a "hint" even mean? Clearly, this is far from a hint. It's an answer. The series $\displaystyle\sum_{n=1}^\infty\frac{1}{n^2-a^2}$ is very typical and common. $\endgroup$ Jul 18, 2016 at 14:51
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    $\begingroup$ Nice to see you around Anastasiya-Romanova :) $\endgroup$
    – FDP
    Jul 18, 2016 at 15:00
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The identity: $$ \sum_{k\geq 1}\frac{\zeta(2k)}{k}x^{2k} = \log\left(\frac{\pi x}{\sin (\pi x)}\right) \tag{1}$$ holds for any $x$ such that $|x|<1$ by the Weierstrass product for the sine function.
By setting $x=\frac{1}{4}$ we recover the wanted result.

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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\color{#f00}{% 4\int_{0}^{\infty}{\sinh^{2}\pars{x/8} \over x\pars{\expo{x} - 1}}\,\dd x} = 4\int_{0}^{\infty}{x \over \expo{x} - 1}\, {\sinh\pars{x/8} \over x}\,{\sinh\pars{x/8} \over x}\,\dd x \\[4mm] = &\ 4\int_{0}^{\infty}{x \over \expo{x} - 1}\,\half\int_{-1/8}^{1/8}\expo{kx}\,\dd k \,\half\int_{-1/8}^{1/8}\expo{qx}\,\dd q\,\dd x = \int_{-1/8}^{1/8}\int_{-1/8}^{1/8}\int_{0}^{\infty} {x\expo{-\pars{1 - k - q}x} \over 1 - \expo{-x}}\,\dd x\,\dd k\,\dd q \\[4mm] = & \int_{-1/8}^{1/8}\,\,\sum_{n = 0}^{\infty}\,\,\int_{-1/8}^{1/8}\int_{0}^{\infty} x\expo{-\pars{1 - k - q + n}x}\,\,\,\,\dd x\,\dd k\,\dd q = \int_{-1/8}^{1/8}\,\,\sum_{n = 0}^{\infty}\,\,\int_{-1/8}^{1/8} {1 \over \pars{1 - k - q + n}^{\,2}}\,\dd k\,\dd q \\[4mm] = &\ {1 \over 4}\int_{-1/8}^{1/8}\,\,\sum_{n = 0}^{\infty} {1 \over \pars{n + 7/8 - q}\pars{n + 9/8 - q}}\,\dd q \\[4mm] = &\ {1 \over 4}\int_{-1/8}^{1/8}\pars{-4} \bracks{\Psi\pars{{7 \over 8} - q} - \Psi\pars{{9 \over 8} - q}}\,\dd q = \left.\vphantom{\huge A^{A}} \ln\pars{\Gamma\pars{7/8 - q} \over \Gamma\pars{9/8 - q}} \right\vert_{\ -1/8}^{\ 1/8} \\[4mm] = &\ \ln\pars{{\Gamma\pars{3/4} \over \Gamma\pars{1}}\, {\Gamma\pars{5/4} \over \Gamma\pars{1}}} = \ln\pars{\Gamma\pars{3 \over 4}\,{1 \over 4}\,\Gamma\pars{1 \over 4}} = \ln\pars{{1 \over 4}\,{\pi \over \sin\pars{\pi/4}}} = \ln\pars{2^{-3/2}\,\,\pi} \\[4mm] = &\ \color{#f00}{\ln\pars{\pi} - {3 \over 2}\,\ln\pars{2}} \end{align}

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