Integral of the day VI

Integral involving the Apéry's constant $\zeta(3)$

Today we show the solution to this integral posted by @sounansya_29 on Twitter \[ \int_{0}^{\infty} \frac{x\ln(\sinh x)}{\sinh x \cosh x} dx = -\frac{7}{16}\zeta(3)\]
The solution involves the integral representation of the Dirichlet lambda function as well as some known results of Euler sums.

Proof

\begin{align*} I=\int_{0}^{\infty} \frac{x\ln(\sinh x)}{\sinh x \cosh x} dx =& 2\int_{0}^{\infty} \frac{x\ln(\sinh(x))}{\sinh(2x)} \quad \left( \sinh x \cosh x = \frac{1}{2}\sinh(2x)\right)\\ =& 2\int_{0}^{\infty} \frac{x\ln\left(\frac{1}{2}(e^{x}-e^{-x})\right)}{\sinh(2x)}\\ =& -2\ln(2)\int_{0}^{\infty} \frac{x}{\sinh(2x)}dx + 2\int_{0}^{\infty} \frac{x\ln(e^{x}(1-e^{-2x}))}{\sinh(2x)}dx\\ =& -2\ln(2)\underbrace{\int_{0}^{\infty} \frac{x}{\sinh(2x)}dx}_{I_{1}} +2\underbrace{\int_{0}^{\infty} \frac{x^2}{\sinh(2x)}dx}_{I_{2}}+ 2\underbrace{\int_{0}^{\infty} \frac{x\ln(1-e^{-2x})}{\sinh(2x)}dx}_{I_{3}}\\ \end{align*} Recall the beautiful integral representation of the Dirichlet lambda function: \[ \lambda(v) = \frac{1}{2\Gamma(v)} = \int_{0}^{\infty} \frac{t^{v-1}}{\sinh(t)}dt \] Therefore, for $I_{1}, I_{2}$ we have: \[I_{1} = \int_{0}^{\infty} \frac{x}{\sinh(2x)}dx = \frac{1}{4}\int_{0}^{\infty} \frac{w}{\sinh(w)} dw = \frac{1}{2}\Gamma(2) \lambda(2) = \frac{\pi^2}{16} \quad (w \mapsto 2x) \] \[I_{2} = \int_{0}^{\infty} \frac{x^2}{\sinh(2x)}dx = \frac{1}{8}\int_{0}^{\infty} \frac{w^2}{\sinh(w)} dw = \frac{1}{4}\Gamma(3) \lambda(3) = \frac{7\zeta(3)}{16} \quad (w \mapsto 2x) \]
For $\displaystyle I_{3}$ we can split the integral in two:
\begin{align*} I_{3} = \int_{0}^{\infty} \frac{x\ln(1-e^{-2x})}{\sinh(2x)}dx =& 2\int_{0}^{\infty} \frac{x\ln(1-e^{-2x})}{e^{2x}-e^{-2x}}dx\\ =& -\frac{1}{2}\int_{0}^{1} \frac{\ln(w) \ln(1-w)}{(1-w)(1+w)}dw \quad (w \mapsto e^{-2x})\\ =& -\frac{1}{4}\underbrace{\int_{0}^{1} \frac{\ln(w) \ln(1-w)}{(1+w)}dw}_{J_{1}} - \frac{1}{4}\underbrace{\int_{0}^{1} \frac{\ln(w) \ln(1-w)}{(1-w)}dw}_{J_{2}} \end{align*} \[ J_{1} = \int_{0}^{1} \frac{\ln(w) \ln(1-w)}{(1+w)}dw \stackrel{IBP}{=} \underbrace{\ln(w)\ln(w+1)\ln(1-w)\Big|_{0}^{1}}_{=0} - \underbrace{\int_{0}^{1} \frac{\ln(w+1)\ln(1-w)}{w} dw}_{K_{1}} + \underbrace{\int_{0}^{1}\frac{\ln(w)\ln(w+1)}{1-w}dw}_{K_{2}} \] \begin{align*} K_{1}=\int_{0}^{1}\frac{\ln(1-x)\ln(1+x)}{x} dx= & \int_{0}^{1}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}x^n}{n}\frac{\ln(1-x)}{x} dx\\ =&\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\int_{0}^{1} x^{n-1}\ln(1-x) dx\\ =&\sum \frac{(-1)^{n+1}}{n}\int_{0}^{1}\sum_{n=1}^{\infty} x^{n-1} \left[\frac{d}{dt}\Big|_{t=0+} (1-x)^t\right] dx\\ =& \frac{d}{dt}\Big|_{t=0+} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\int_{0}^{1}\sum x^{n-1}(1-x)^t dx\\ =&\frac{d}{dt}\Big|_{t=0+} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}B(n,t+1)\\ =& \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}\left(\psi^{(0)}(n+1)+\gamma\right)B(n,1)\\ =& \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^2}H_{n}\\ \end{align*} \begin{align*} K_{2} = \int_{0}^{1}\frac{\ln(w)\ln(w+1)}{1-w}dw = &\int_{0}^{1} \frac{\ln(1-s)\ln(2-s)}{s}ds \\ =& \int_{0}^{1} \frac{\ln(1-s)\ln\left(2\left(1-\frac{s}{2}\right)\right)}{s}ds\\ =&\ln(2) \int_{0}^{1} \frac{\ln(1-s)}{s}ds + \int_{0}^{1} \frac{\ln(1-s)\ln\left(1-\frac{s}{2}\right)}{s}ds \\ =& -\ln(2) \int_{1}^{0} \frac{\ln(1-s)}{s}ds - \int_{0}^{1} \frac{\ln(1-s)}{s} \sum_{n=1}^{\infty} \frac{s^{n}}{2^nn}ds \\ =& -\ln(2)\operatorname{Li}_{2}(1) - \sum_{n=1}^{\infty} \frac{1}{2^nn}\int_{0}^{1} s^{n-1}\ln(1-s)ds \\ =& -\frac{\ln(2)\pi^2}{6} - \sum_{n=1}^{\infty} \frac{1}{2^nn}\int_{0}^{1} s^{n-1} \left(\frac{d}{dt}\Big|_{t=0+} (1-s)^t\right) ds\\ =& -\frac{\ln(2)\pi^2}{6} - \frac{d}{dt}\Big|_{t=0+} \sum_{n=1}^{\infty} \frac{1}{2^nn}\int_{0}^{1} s^{n-1} (1-s)^t ds\\ =& -\frac{\ln(2)\pi^2}{6} - \frac{d}{dt}\Big|_{t=0+} \sum_{n=1}^{\infty} \frac{1}{2^nn}B(n,t+1)\\ =& -\frac{\ln(2)\pi^2}{6} + \sum_{n=1}^{\infty} \frac{H_{n}}{2^nn^2}\\ \end{align*} \begin{align*} J_{2} = \int_{0}^{1} \frac{\ln(1-w)\ln(w)}{1-w}dw =& \underbrace{-\frac{1}{2} \ln(w)\ln^2(1-w)\Big|_{0}^{1}}_{=0} +\frac{1}{2}\int_{0}^{1} \frac{\ln^2(1-w)}{w}dw \\ =& \frac{1}{2}\int_{0}^{1} \frac{\ln^2(1-w)}{w}dw \\ =& -\frac{1}{2} \int_{0}^{1} \sum_{n=1}^{\infty} \frac{w^n}{n} \frac{\ln(1-w)}{w} dw \\ =& - \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{1} w^{n-1} \left(\frac{d}{dt}\Big|_{t=0+} (1-w)^t \right) dw \\ =&\frac{d}{dt}\Big|_{t=0+} - \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{1} w^{n-1}(1-w)^t dw\\ =& \frac{d}{dt}\Big|_{t=0+} - \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n}B(n,t+1)\\ =& \frac{1}{2} \sum_{n=1}^{\infty}\frac{H_{n}}{n^2} \end{align*} From the generating function (I will add the proof of the generating functions to the appendix later) \[ \sum_{j=1}^{\infty} \frac{H_{j}(1-z)^j}{j^2} = \frac{1}{2}\ln^2(z)\ln(1-z)+\operatorname{Li}_{2}(z)\ln(z) - \operatorname{Li}_{3}(z) + \operatorname{Li}_{3}(1-z)\] If we put $ z = \frac{1}{2}$ we have: \[\sum_{n=1}^{\infty} \frac{H_{n}}{2^nn^2} = \zeta(3) - \frac{\pi^2\ln(2)}{12} \] If $z \to 0$ \[\sum_{n=1}^{\infty} \frac{H_{n}}{n^2} = \zeta(3) \] while from the generating function \[\sum_{j=1}^{\infty} \frac{H_{j}(z)^j}{j^2} = \frac{1}{3}\ln^3(1-z) +\ln(1-z)\operatorname{Li}_{2}\left(\frac{z}{z-1}\right) + \operatorname{Li}_{3}\left(\frac{z}{z-1}\right) +2 \operatorname{Li}_{3}(z) \quad z\in\left[-1,\frac{1}{2}\right]\] If we put $\displaystyle z = -1$ we have \[\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^2}H_{n} = -\frac{5}{8}\zeta(3)\] Therefore \[ J_{1} = -K_{1} + K_{2} = -\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^2}H_{n} + -\frac{\ln(2)\pi^2}{6} + \sum_{n=1}^{\infty} \frac{H_{n}}{2^nn^2} = \frac{13}{8}\zeta(3) - \frac{\pi^2\ln(2)}{4} \] \[J_{2} = \frac{1}{2} \sum_{n=1}^{\infty}\frac{H_{n}}{n^2} = \zeta(3) \] \[I_{3} = -\frac{1}{4}J_{1} -\frac{1}{4}J_{2} = -\frac{21}{32}\zeta(3)+\frac{\pi^2\ln(2)}{16} \] \[ I = -2\ln(2)I_{1} + 2I_{2} +2I_{3} = -2\ln(2)\frac{\pi^2}{16} + 2\frac{7\zeta(3)}{16} -\frac{42}{32}\zeta(3)+\frac{\pi^2\ln(2)}{8} = -\frac{7}{16}\zeta(3) \] Hence, we can conclude \[\boxed{ \int_{0}^{\infty} \frac{x\ln(\sinh x)}{\sinh x \cosh x} dx = -\frac{7}{16}\zeta(3)}\]