Evaluation of $4\int_0^{+\infty} \frac{\left(\sinh\left(\frac{x}{8}\right)\right)^2}{x(e^x-1)}dx$

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}