Closed-form of $\int\limits_0^1\left(\frac{\left(x^2+1\right)\arcsin(x)}{\sqrt{1-x^2}}+2x\ln\left(x^2+1\right)\right)\frac{\ln x}{x^3+x}\,dx$

I've conjectured the following closed-form: $$ I = \int\limits_0^1\left(\frac{\left(x^2+1\right)\arcsin(x)}{\sqrt{1-x^2}}+2x\ln\left(x^2+1\right)\right)\frac{\ln x}{x^3+x}\,dx = -2\,G\,\ln2, $$ where $G$ is Catalan's constant. Numerically $$ I \approx -1.2697979381877088371491554851603117320986537271546606092465\dots$$ How to prove it?

Solution 1:

Quoting this famous question, we have that: $$ \sum_{n\geq 1}\frac{2^{2n} x^{2n}}{n^2\binom{2n}{n}}=2\,\arcsin^2(x)\tag{1}$$ hence: $$\begin{eqnarray*} \int\limits_{0}^{1}\frac{\arcsin(x)}{\sqrt{1-x^2}}\cdot\frac{\log x}{x}\,dx &=& -\frac{1}{4}\sum_{n\geq 1}\frac{4^n}{n(2n-1)^2\binom{2n}{n}}\\&=&-\frac{1}{2}\sum_{n\geq 1}\frac{4^n\,B(n,n)}{(2n)^2(2n-1)^2}\\&=&-2\int\limits_{0}^{1}\sum_{n\geq 1}\frac{\left(4x(1-x)\right)^{n-1}}{(2n)^2(2n-1)^2}\,dx.\tag{2}\end{eqnarray*}$$ On the other hand, $ \log^2(1-x) = \sum_{n\geq 1}\frac{2H_{n-1}}{n}x^n $ implies: $$ \log^2(1+x^2) = \sum_{n\geq 1}\frac{2(-1)^n H_{n-1}}{n}x^{2n}\tag{3} $$ hence: $$ \int\limits_{0}^{1}\frac{\log(1+x^2)\log(x)}{(1+x^2)x}\,dx = \frac{1}{4}\sum_{n\geq 1}\frac{(-1)^n H_{n-1}}{n^2}=-\frac{5}{32}\zeta(3)\tag{4}$$ and the problem boils down to evaluating $(2)$. We have:

$$\begin{eqnarray*} \sum_{n\geq 1}\frac{y^{n-1}}{(2n)^2(2n-1)^2}&=&\sum_{n\geq 1}\left(\frac{1}{(2n)^2}+\frac{1}{(2n-1)^2}+\frac{1}{n}-\frac{2}{2n-1}\right)y^{n-1}\tag{5}\\&=&\frac{\text{Li}_2(y)}{4y}-\frac{\log(1-y)}{y}-\frac{2\,\text{arctanh}(\sqrt{y})}{\sqrt{y}}+\frac{\text{Li}_2(\sqrt{y})-\text{Li}_2(-\sqrt{y})}{2\sqrt{y}}\end{eqnarray*}$$ and, according to Mathematica: $$ \int\limits_{0}^{1}\frac{\text{Li}_2(4x(1-x))}{x(1-x)}\,dx=2\pi^2\log 2-7\zeta(3), $$ $$ \int\limits_{0}^{1}\frac{\log(1-4x(1-x))}{x(1-x)}\,dx = -\pi^2,$$ $$ \int\limits_{0}^{\pi/2}\text{arctanh}(\sin \theta)\,d\theta = 2K,\tag{6} $$ and the last two integrals left to compute are: $$ I_+ = \int\limits_{0}^{\pi/2}\text{Li}_2(\sin\theta)\,d\theta, \qquad I_-=\int\limits_{0}^{\pi/2}\text{Li}_2(-\sin\theta)\,d\theta.$$ They can be computed through integration by parts, then by exploiting the Fourier series of $t\cot t$ and $\log(\sin t),\log(\cos t)$. For the latter, see this question.

Footnote: I still need to complete this answer by properly explaining the last part and proving the identities kindly provided by Mathematica. Since it is quite a lot to typeset, I leave this answer here even if not complete: maybe someone finds one or more shortcuts.

Solution 2:

Apply the obvious substitution $x\mapsto\sin{x}$ to the first integral $$I=\int^\frac{\pi}{2}\limits_0\frac{x\ln(\sin{x})}{\sin{x}}\ {\rm d}x+\int\limits^1_0\frac{2\ln{x}\ln(1+x^2)}{1+x^2}\ {\rm d}x$$ The latter integral has been addressed here and is equivalent to \begin{align} \int\limits^1_0\frac{2\ln{x}\ln(1+x^2)}{1+x^2}\ {\rm d}x=4\Im{\rm Li}_3(1-i)-2\mathbf{G}\ln{2}+\frac{3\pi^3}{16}+\frac{\pi}{4}\ln^2{2} \end{align} As for the first integral, \begin{align} \int\limits^\frac{\pi}{2}_0\frac{x\ln(\sin{x})}{\sin{x}}\ {\rm d}x &=2\int\limits^1_0\frac{\arctan{x}\ln\left(\frac{2x}{1+x^2}\right)}{x}\ {\rm d}x\\ &=2{\rm Ti}_2(1)\ln{2}+2\int\limits^1_0\frac{\arctan{x}\ln{x}}{x}\ {\rm d}x-2\int\limits^1_0\frac{\arctan{x}\ln(1+x^2)}{x}\ {\rm d}x\\ &=2\mathbf{G}\ln{2}-2\sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)^3}+2\int\limits^1_0\frac{\ln{x}\ln(1+x^2)}{1+x^2}\ {\rm d}x+4\int\limits^1_0\frac{x\arctan{x}\ln{x}}{1+x^2}\ {\rm d}x \end{align} and the integral \begin{align} 4\int\limits^1_0\frac{x\arctan{x}\ln{x}}{1+x^2}\ {\rm d}x=-8\Im{\rm Li}_3(1-i)-\frac{5\pi^3}{16}-\frac{\pi}{2}\ln^2{2} \end{align} has also been established in the link above. (Both integrals were covered in the evaluation of $\mathscr{J}_2$.) Hence the closed form is indeed \begin{align} I=-2\mathbf{G}\ln{2} \end{align}