TITLE: A rare integral involving $\operatorname{Li}_2$
QUESTION [11 upvotes]: A rare but interesting integral problem:
$$\int_{0}^{1} 
\frac{\operatorname{Li}_2(-x)-
\operatorname{Li}_2(1-x)+\ln(x)\ln(1+x)+\pi x\ln(1+x)
-\pi x\ln(x)}{1+x^2}\frac{\text{d}x}{\sqrt{1-x^2} } 
=\frac{\pi^3}{48\sqrt{2} }.$$
Where $\operatorname{Li}_2$ is dilogarithm.
The integral without the factor $\frac{1}{\sqrt{1-x^2} }$ is much easier(it can be expressed using $\operatorname{Li}_3$,$\ln(2)$,Catalan's constant and $\pi$).
However, the same idea dosen't work for this one. Any suggestion will be much appreciated.

REPLY [17 votes]: We begin by separating the integral into three parts:
$$I = \pi \color{red}{\overbrace{\int_{0}^{1} \frac{x \ln (1+x)}{(1+x^2) \sqrt{1-x^2}} \, dx}^J} - \pi \color{blue}{\overbrace{\int_{0}^{1} \frac{ x \ln (x)}{(1+x^2) \sqrt{1-x^2}} \, dx}^K} + \color{purple}{\underbrace{\int_{0}^{1}\frac{\operatorname{Li}_2(-x)-\operatorname{Li}_2(1-x)+\ln(x)\ln(1+x)}{(x^2+1)\sqrt{1-x^2}} \, dx}_L}$$
We begin by evaluating $\color{red}{J}$:
$$J = \int_{0}^{1}\frac{x\ln(x+1)}{(x^2+1)\sqrt{1-x^2}} \, dx$$
Consider
$$\int_{0}^{1}\frac{x(\ln(x+1)+\ln(1-x))}{(x^2+1)\sqrt{1-x^2}} \, dx=-\frac{2}{\sqrt{2}}\int_{0}^{1}\frac{x}{1-x^2}\text{arctanh}\left(\frac{\sqrt{1-x^2}}{\sqrt2}\right) \, dx$$
$$=-\frac{2}{\sqrt{2}}\int_{0}^{1}\frac{1}{u}\text{arctanh}\left(\frac{u}{\sqrt2}\right)\,du = -\frac{2}{\sqrt2}\left(-\frac{\pi^2}{48}+\frac{\ln^2(2)}{8}+\operatorname{Li}_2\left(\frac{1}{\sqrt2}\right)\right)$$
Consider now
$$\int_{0}^{1}\frac{x(\ln(x+1)-\ln(1-x))}{(x^2+1)\sqrt{1-x^2}} \,dx=\frac{2}{\sqrt{2}}\int_{0}^{1}\frac{1}{1-x^2}\text{arctanh}\left(\frac{\sqrt{1-x^2}}{\sqrt2}\right)\,dx$$
$$=\frac{2}{\sqrt{2}}\int_{0}^{1}\frac{1}{u\sqrt{1-u^2}}\text{arctanh}\left(\frac{u}{\sqrt2}\right) \,du=\frac{\pi^2}{4\sqrt2}$$
Thus summing the two we have
$$\color{red}{\implies J =\frac{\pi^2}{8\sqrt2}-\frac{1}{\sqrt2}\left(-\frac{\pi^2}{48}+\frac{\ln^2(2)}{8}+\operatorname{Li}_2\left(\frac{1}{\sqrt2}\right)\right)}$$
Now we evaluate $\color{blue}{K}$:
$$K = \int_{0}^{1} \frac{x \ln (x)}{(1+x^2)\sqrt{1-x^2}} \, dx$$
$$\int_{0}^{1}\frac{x \ln(x)}{\sqrt{1-x^2}(1+x^2)}\,dx=\frac{1}{\sqrt{2}}\int_{0}^{1}\frac{1}{x}\left(\text{arctanh}\left(\frac{\sqrt{1-x^2}}{\sqrt2}\right)-\text{arctanh}\left(\frac{1}{\sqrt2}\right)\right)\,dx$$
$$=\frac{1}{\sqrt{2}}\int_{0}^{1}\frac{u}{1-u^2}\left(\text{arctanh}\left(\frac{u}{\sqrt2}\right)-\text{arctanh}\left(\frac{1}{\sqrt2}\right)\right)\,du$$
We shall now approach this with differentiation under the integral sign.
Define
$$f(b)=\frac{1}{\sqrt{2}}\int_{0}^{1}\frac{u}{1-u^2}\left(\text{arctanh}\left(\frac{u}{b}\right)-\text{arctanh}\left(\frac{1}{b}\right)\right)\,du$$
$$f'(b)=\frac{1}{2 \sqrt{2}}\left(\frac{b(\ln(b+1)-\ln(b-1))-2\ln(2)}{b^2-1}\right)$$
Notice that $\lim_{b \to \infty} f(b) = 0 \implies f(\sqrt{2})-f(\infty)=K$
$$\color{blue}{\implies K=-\frac{1}{4\sqrt{2}}\left(\frac{\pi^2}{3}+2\operatorname{Li}_2\left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)+\ln^2(2)-\text{arcsinh}(1)^2-2\ln(2)\text{arcsinh}(1)\right)}$$
Now we evaluate $\color{purple}{L}$:
$$L = \int_{0}^{1}\frac{\operatorname{Li}_2(-x)-\operatorname{Li}_2(1-x)+\ln(x)\ln(1+x)}{(x^2+1)\sqrt{1-x^2}} \,dx \\= -\frac{\pi^3}{24\sqrt2}+\frac{2}{\sqrt{2}}\int_{0}^{1}\frac{x\ln(x)}{1-x^2}\text{arctan}\left(\frac{\sqrt{2}x}{\sqrt{1-x^2}}\right) \,dx$$
$$=-\frac{\pi^3}{24 \sqrt{2}}+\frac{2}{\sqrt{2}}\int_{0}^{\infty}\frac{u}{1+u^2}\ln\left(\frac{u}{\sqrt{u^2+1}}\right)\text{arctan}\left(\sqrt{2} u \right)\,du$$
$$=-\frac{\pi^3}{12\sqrt{2}} + \frac{2}{\sqrt{2}}\int_{0}^{\infty}\frac{u}{1+u^2}\ln\left(\frac{u}{\sqrt{u^2+1}}\right)\left(\text{arctan}\left(\sqrt{2}u\right)-\frac{\pi}{2}\right)\,du$$
Now separate the logarithm and consider the two integrals
$$\color{green}{A=\int_{0}^{\infty}\frac{u}{1+u^2}\ln\left(u^2+1\right)\left(\text{arctan}\left(u\sqrt2\right)-\frac{\pi}{2}\right) \, du}$$
$$\color{brown}{B=\int_{0}^{\infty}\frac{u}{1+u^2}\ln\left(u\right)\left(\text{arctan}\left(u\sqrt2\right)-\frac{\pi}{2}\right) \, du}$$
such that $\color{purple}{L} = -\frac{\pi^3}{12 \sqrt{2}} + \frac{2}{\sqrt{2}}\color{brown}{B}-\frac{1}{\sqrt{2}} \color{green}{A}$.
We begin by evaluating $\color{green}{A}$ via differentiation under the integral sign:
$$A=\int_{0}^{\infty}\frac{u}{1+u^2}\ln\left(u^2+1\right)\left(\text{arctan}\left(u\sqrt2\right)-\frac{\pi}{2}\right) \, du$$
Recall
$$\ln(u^2+1)=2\int_{0}^{\infty}\frac{1-\cos(ux)}{x}e^{-x}\,dx$$
Define
$$g(b) = \int_{0}^{\infty}\frac{u}{1+u^2}\ln\left(u^2+1\right)\left(\text{arctan}\left(\frac{u}{b}\right)-\frac{\pi}{2}\right)\,du$$
$$g'(b) = -2\int_{0}^{\infty}\frac{u^2}{(1+u^2)(b^2+u^2)}\int_{0}^{\infty}\frac{1-\cos(ux)}{x}e^{-x}\,dx\,du$$
$$=-\frac{\pi}{(1-b^2)}\int_{0}^{\infty}\frac{e^{-x}}{x}\left(1-e^{-x}-b+be^{-xb}\right)\,dx =\frac{\pi b\ln(1+b)-\pi\ln(2)}{1-b^2}$$
Notice $\lim_{b \to 0} g(b) = 0 \implies g \left(\frac{1}{\sqrt{2}}\right) - g(0) = A$
$$\color{green}{\implies A = -\frac{\pi^3}{24}-\pi \, \text{arccoth}(\sqrt{2}) \,\ln(2)-\frac{\pi}{2}\text{arctanh}\left(\frac{1}{7}+\frac{2\sqrt2}{7}\right)\ln\left(\frac{2+\sqrt2}{16}\right)+\frac{\pi}{2}\operatorname{Li}_2\left(\frac{2-\sqrt2}{4}\right)}$$
Similarly, we evaluate $\color{brown}{B}$ via differentiation under the integral sign:
$$B=\int_{0}^{\infty}\frac{u}{1+u^2}\ln\left(u\right)\left(\text{arctan}\left(u\sqrt2\right)-\frac{\pi}{2}\right) \, du$$
Define
$$h(b) = \int_{0}^{\infty}\frac{u}{1+u^2}\ln\left(u\right)\left(\text{arctan}\left(\frac{u}{b}\right)-\frac{\pi}{2}\right) \, du$$
$$h'(b) = -\int_{0}^{\infty}\frac{u^2\ln(u)}{(u^2+1)(b^2+u^2)} \, du = \frac{\pi \, b \ln(b)}{2(1-b^2)}$$
Notice $\lim_{b \to 0} h(b) = 0 \implies h \left(\frac{1}{\sqrt{2}}\right) - h(0) = B$
$$\color{brown}{\implies B = -\frac{\pi^3}{96}-\frac{\pi \ln^2(2)}{16}}$$
Thus we have
$$\color{purple}{L} = -\frac{\pi^3}{12 \sqrt{2}} + \frac{2}{\sqrt{2}} \color{brown}{\left(-\frac{\pi^3}{96}-\frac{\pi \ln^2(2)}{16}\right)} - \frac{1}{\sqrt{2}} \color{green}{\left( 
-\frac{\pi^3}{24}-\pi \, \text{arccoth}(\sqrt{2}) \,\ln(2)-\frac{\pi}{2}\text{arctanh}\left(\frac{1}{7}+\frac{2\sqrt2}{7}\right)\ln\left(\frac{2+\sqrt2}{16}\right)+\frac{\pi}{2}\operatorname{Li}_2\left(\frac{2-\sqrt2}{4}\right) \right)}$$
$$\color{purple}{\implies L = -\frac{\pi\operatorname{Li}_2\left(\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)}{2\sqrt{2}}-\frac{\pi^3}{16\sqrt{2}}-\frac{\pi\ln^2(2)}{8\sqrt{2}}-\frac{\pi\ln\left(16-8\sqrt{2}\right)\, \text{arctanh}\left(\frac{1}{7}+\frac{2\sqrt{2}}{7}\right)}{2\sqrt{2}}+\frac{\pi\ln(2)\,\text{arccoth}\left(\sqrt{2}\right)}{\sqrt{2}}}$$
$$\color{purple}{\implies L = -\frac{\pi\operatorname{Li}_2\left(\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)}{2\sqrt{2}}-\frac{\pi^3}{16\sqrt{2}}-\frac{7\pi\ln^2(2)}{8\sqrt{2}}+\frac{\pi\ln^2\left(2-\sqrt{2}\right)}{4\sqrt{2}}+\frac{\pi\ln(2)\ln\left(2+\sqrt{2}\right)}{2\sqrt{2}}}$$
Thus finally, we have
$$I = \pi \color{red}{\left(\frac{\pi^2}{8\sqrt2}-\frac{1}{\sqrt2}\left(-\frac{\pi^2}{48}+\frac{\ln^2(2)}{8}+\operatorname{Li}_2\left(\frac{1}{\sqrt2}\right)\right)\right)} - \pi \color{blue}{\left( -\frac{1}{4\sqrt{2}}\left(\frac{\pi^2}{3}+2\operatorname{Li}_2\left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)+\ln^2(2)-\text{arcsinh}(1)^2-2\ln(2)\text{arcsinh}(1)\right) \right)} + \color{purple}{\left( -\frac{\pi\operatorname{Li}_2\left(\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)}{2\sqrt{2}}-\frac{\pi^3}{16\sqrt{2}}-\frac{7\pi\ln^2(2)}{8\sqrt{2}}+\frac{\pi\ln^2\left(2-\sqrt{2}\right)}{4\sqrt{2}}+\frac{\pi\ln(2)\ln\left(2+\sqrt{2}\right)}{2\sqrt{2}} \right) }$$
After some simplification
$$\implies I = -\frac{\pi\operatorname{Li}_2\left(\frac{1}{\sqrt{2}}\right)}{\sqrt{2}}+\frac{\pi\operatorname{Li}_2\left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)}{2\sqrt{2}}-\frac{\pi\operatorname{Li}_2\left(\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)}{2\sqrt{2}}+\frac{\pi^3}{6\sqrt{2}}-\frac{3\pi\ln^2(2)}{4\sqrt{2}}+\frac{\pi\ln^2\left(2-\sqrt{2}\right)}{4\sqrt{2}}-\frac{\pi\ln^2\left(1+\sqrt{2}\right)}{4\sqrt{2}}-\frac{\pi\ln(2)\ln\left(1+\sqrt{2}\right)}{2\sqrt{2}}+\frac{\pi\ln(2)\ln\left(2+\sqrt{2}\right)}{2\sqrt{2}}$$
After some dilogarithm identities
$$I = \frac{\pi^3}{48 \sqrt{2}}$$
$\square$