Integral of the day X

Integral involving the Glaisher–Kinkelin constant

We show the proof of this weird integral posted by @integralsbot \[\int_{0}^{1} \frac{\ln(x)}{1-x^2} \ln\left(\ln\left(\frac{1+x}{1-x}\right)\right)dx = \frac{\pi^2}{24}\ln\left(\frac{A^{36}}{16\pi^3}\right)\] where $A$ is the Glaisher–Kinkelin constant

Proof

Recall the logarithmic representation of $\operatorname{arctanh}(x)$: \[\operatorname{arctanh}(x) = \ln\left(\sqrt{\frac{1+x}{1-x}}\right)\] Hence \begin{align*} I=\int_{0}^{1} \frac{\ln(x)}{1-x^2} \ln\left(\ln\left(\frac{1+x}{1-x}\right)\right)dx =& \int_{0}^{1} \frac{\ln(x)}{1-x^2} \ln\left(2\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\right)dx\\ =&\ln(2)\int_{0}^{1} \frac{\ln(x)}{1-x^2}dx + \int_{0}^{1} \frac{\ln(x)}{1-x^2} \ln\left(\ln\left(\sqrt{\frac{1+x}{1-x}}\right)\right)dx\\ =&\ln(2) \underbrace{\int_{0}^{1} \frac{\ln(x)}{1-x^2}dx}_{J} + \underbrace{\int_{0}^{1} \frac{\ln(x)}{1-x^2} \ln\left(\operatorname{arctanh}(x)\right)dx}_{K}\\ \end{align*} \[ J = \int_{0}^{1} \frac{\ln(x)}{1-x^2}dx = \frac{1}{2}\underbrace{\int_{0}^{1}\frac{\ln(x)}{x+1}dx}_{L} - \frac{1}{2}\underbrace{\int_{0}^{1}\frac{\ln(x)}{1-x}dx}_{M} \] \[ L = \int_{0}^{1}\frac{\ln(x)}{x+1}dx \stackrel{IBP}{=} \ln(x)\ln(x+1)\Big|_{0}^{1} + \int_{-1}^{0} \frac{\ln(1-x)}{x}dx = \operatorname{Li}_{2}(-1) = -\frac{\pi^2}{12} \] \[ M = \int_{0}^{1}\frac{\ln(x)}{1-x}dx = -\int_{1}^{0} \frac{\ln(1-w)}{w} dw = -\operatorname{Li}_{2}(1) = -\frac{\pi^2}{6} \] \[ \therefore J = \frac{1}{2}L+\frac{1}{2}M = -\frac{\pi^2}{24} -\frac{\pi^2}{12} = -\frac{\pi^2}{8} \] Now, for $K$: \begin{align*} K = \int_{0}^{1} \frac{\ln(x)}{1-x^2} \ln\left(\operatorname{arctanh}(x)\right)dx =& \int_{0}^{\infty} \ln\left(\operatorname{tanh}(w)\right) \ln(w)dw \quad ( w \mapsto \operatorname{arctanh}(x))\\ =&\int_{0}^{\infty} \ln\left(\frac{e^{w}-e^{-w}}{e^{w}+e^{-w}}\right) \ln(w)dw \\ =&\int_{0}^{\infty} \ln\left(\frac{1-e^{-2w}}{1+e^{-2w}}\right) \ln(w)dw \\ =&\underbrace{\int_{0}^{\infty} \ln\left(1-e^{-2w}\right) \ln(w)dw}_{C} - \underbrace{\int_{0}^{\infty} \ln\left(1+e^{-2w}\right) \ln(w)dw }_{D} \end{align*} \begin{align*} C = \int_{0}^{\infty} \ln\left(1-e^{-2w}\right) \ln(w)dw =& -\int_{0}^{\infty} \sum_{n=1}^{\infty}\frac{ e^{-2nw}}{n} \ln(w)dw\\ =& -\sum_{n=1}^{\infty} \frac{1}{n}\int_{0}^{\infty} e^{-2nw} \ln(w)dw\\ =& -\sum_{n=1}^{\infty} \frac{1}{n}\int_{0}^{\infty} e^{-2nw} \left(\frac{d}{dt}\Big|_{t=0+} w^t \right)dw\\ =&\frac{d}{dt}\Big|_{t=0+} -\sum_{n=1}^{\infty} \frac{1}{n}\int_{0}^{\infty} e^{-2nw} w^tdw\\ =&\frac{d}{dt}\Big|_{t=0+} -\sum_{n=1}^{\infty} \frac{1}{n^{t+2}2^{t+1}}\int_{0}^{\infty} e^{-s} s^tds\\ =&\frac{d}{dt}\Big|_{t=0+} -\sum_{n=1}^{\infty} \frac{1}{2^{t+1}n^{t+2}}\Gamma(t+1)\\ =& -\sum_{n=1}^{\infty} \left[\frac{1}{2n^2}\frac{d}{dt}\Big|_{t=0+}\frac{1}{(2n)^{t}} \Gamma(t+1)\right]\\ =& \sum_{n=1}^{\infty}\frac{\ln(2n)+\gamma}{2n^2}\\ =& \frac{\ln(2)}{2} \sum_{n=1}^{\infty}\frac{1}{n^2} + \frac{1}{2} \sum_{n=1}^{\infty}\frac{\ln(n)}{n^2} + \frac{\gamma}{2} \sum_{n=1}^{\infty}\frac{1}{n^2}\\ =& \frac{\ln(2)}{2} \zeta(2) - \frac{1}{2} \zeta'(2) + \frac{\gamma}{2} \zeta(2)\\ =& \frac{\ln(2)\pi^2}{12} - \frac{1}{2} \zeta'(2) + \frac{\gamma\pi^2}{12} \end{align*} Recall \[ -\zeta'(2) = 2\pi^2 \ln(A) - \gamma\frac{\pi^2}{6} - \ln(2)\frac{\pi^2}{6} -\ln(\pi)\frac{\pi^2}{6} \tag{*} \] whre $A$ is the Glaisher–Kinkelin constant. Hence \[ C = \int_{0}^{\infty} \ln\left(1-e^{-2w}\right) \ln(w)dw = \pi^2\ln(A) - \frac{\pi^2\ln(\pi)}{12} \] In a similar way: \begin{align*} D = \int_{0}^{\infty} \ln\left(1+e^{-2w}\right) \ln(w)dw =& \int_{0}^{\infty} \sum_{n=0}^{\infty} \frac{(-1)^{n+1}e^{-2nw}}{n} \ln(w)dw\\ =& \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{n}\int_{0}^{\infty} e^{-2nw} \ln(w)dw\\ =& \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{n}\int_{0}^{\infty} e^{-2nw} \left(\frac{d}{dt}\Big|_{t=0+} w^t\right)dw\\ =& \frac{d}{dt}\Big|_{t=0+} \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{n}\int_{0}^{\infty} e^{-2nw} w^tdw\\ =&\frac{d}{dt}\Big|_{t=0+} \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{2^{t+1}n^{t+2}}\int_{0}^{\infty} e^{-s} s^{t}ds\\ =&\sum_{n=0}^{\infty} \left[ \frac{(-1)^{n+1}}{2n^2} \frac{d}{dt}\Big|_{t=0+} \frac{\Gamma(t+1)}{2^{t}n^{t}}\right]\\ =&\sum_{n=0}^{\infty} \frac{(-1)^{n}(\ln(2n)+\gamma)}{2n^2}\\ =&\frac{\ln(2)}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n^2} + \frac{1}{2}\sum_{n=0}^{\infty} \frac{(-1)^{n}\ln(n)}{n^2} + \frac{\gamma}{2}\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n^2}\\ =&\frac{-\ln(2)}{2}\eta(2) +\frac{1}{2}\eta'(2) - \frac{\gamma}{2}\eta(2)\\ =&-\frac{\ln(2)\pi^2}{24} + \frac{1}{2}\eta'(2) - \frac{\gamma\pi^2}{24}\\ \end{align*} Using the relation \[ \eta(v) = (1-2^{1-v})\zeta(v)\] and using (*) we have \[\eta'(2) = -\pi^2\ln(A) +\frac{\gamma\pi^2}{12} + \frac{\pi^2\ln(\pi)}{12} + \frac{\pi^2\ln(2)}{6}\] Hence \[ D = \int_{0}^{\infty} \ln\left(1+e^{-2w}\right) \ln(w)dw= -\frac{\pi^2\ln(A)}{2} + \frac{\ln(\pi)\pi^2}{24} + \frac{\ln(2)\pi^2}{24} \] Therefore \[ K = C-D =\pi^2\ln(A) - \frac{\pi^2\ln(\pi)}{12} + \frac{\pi^2\ln(A)}{2} - \frac{\ln(\pi)\pi^2}{24} - \frac{\ln(2)\pi^2}{24} = \frac{3\pi^2}{2} \ln(A)-\frac{\pi^2\ln(\pi)}{8}-\frac{\ln(2)\pi^2}{24} \] Therefore \begin{align*} I = \ln(2)J+K =-\frac{\pi^2\ln(2)}{8} + \frac{3\pi^2}{2} \ln(A)-\frac{\pi^2\ln(\pi)}{8}-\frac{\ln(2)\pi^2}{24} =& \frac{\pi^4}{24}\left[-4\ln(2)+36\ln(A)-3\ln(\pi) \right]\\ =& \frac{\pi^2}{24}\ln\left(\frac{A^{36}}{16\pi^3}\right) \end{align*} Hence, we can conclude \[\boxed{\int_{0}^{1} \frac{\ln(x)}{1-x^2} \ln\left(\ln\left(\frac{1+x}{1-x}\right)\right)dx = \frac{\pi^2}{24}\ln\left(\frac{A^{36}}{16\pi^3}\right)}\]