Prove $\int_0^1\frac{\ln x\ln(1+x)}{1-x}\ dx=\zeta(3)-\frac32\ln2\zeta(2)$

\begin{align}J&=\int_0^1 \frac{\ln x\ln(1+x)}{1-x}\,dx\end{align} Always the same story...

For $x\in [0;1]$ define the function $R$ by,\begin{align}R(x)&=\int_0^x \frac{\ln t}{1-t}\,dt\\ &=\int_0^1 \frac{x\ln(tx)}{1-tx}\,dt\\ J&=\Big[R(x)\ln(1+x)\Big]_0^1-\int_0^1 \frac{R(x)}{(1+x)} dx\\ &=-\zeta(2)\ln 2-\int_0^1 \int_0^1 \frac{x\ln(tx)}{(1-tx)(1+x)}\,dt\,dx\\ &=-\zeta(2)\ln 2-\int_0^1 \left(\int_0^1 \frac{x\ln t}{(1-tx)(1+x)}\,dx\right)\,dt-\int_0^1 \left(\int_0^1 \frac{x\ln x}{(1-tx)(1+x)}\,dt\right)\,dx\\ &=-\zeta(2)\ln 2+\int_0^1\left[\frac{\ln(1-tx)}{t(1+t)}+\frac{\ln(1+x)}{1+t}\right]_{x=0}^{x=1}\ln t\,dt+\int_0^1\left[\frac{\ln(1-tx)}{1+x}\right]_{t=0}^{t=1}\ln x\,dx\\ &=-\zeta(2)\ln 2+\int_0^1 \frac{\ln(1-t)\ln t}{t(1+t)}\,dt+\ln 2\int_0^1 \frac{\ln t}{1+t}\,dt+\int_0^1 \frac{\ln(1-x)\ln x}{1+x}\,dx\\ &=-\zeta(2)\ln 2+\int_0^1 \frac{\ln(1-t)\ln t}{t}\,dt+\ln 2\int_0^1 \frac{\ln t}{1+t}\,dt\\ &=-\zeta(2)\ln 2+\frac{1}{2}\left(\Big[\ln^2 x\ln(1-x)\Big]+\int_0^1 \frac{\ln^2 t }{1-t}\,dt\right)+\ln 2\left(\int_0^1 \frac{\ln t}{1-t}\,dt-\int_0^1 \frac{2t\ln t}{1-t^2}\,dt\right) \end{align} In the last integral perform the change of variable $y=t^2$, \begin{align}J&=-\zeta(2)\ln 2+\frac{1}{2}\int_0^1 \frac{\ln^2 t}{1-t}\,dt+\ln 2\left(\int_0^1 \frac{\ln t}{1-t}\,dt-\frac{1}{2}\int_0^1 \frac{\ln t}{1-t}\,dt\right)\\ &=-\frac{3}{2}\zeta(2)\ln 2+\frac{1}{2}\times 2\zeta(3)\\ &=\boxed{-\frac{3}{2}\zeta(2)\ln 2+\zeta(3)} \end{align} NB: i assume,\begin{align}R(1)&=\int_0^1 \frac{\ln t}{1-t}\,dt\\ &=-\zeta(2)\\ \int_0^1 \frac{\ln^2 t}{1-t}\,dt&=2\zeta(3)\end{align}

Start off with the substitution $x\to \frac{1-x}{1+x}$ to get: $$\require{cancel} I=\int_0^1 \frac{\ln x\ln(1+x)}{1-x}dx=\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{2}{1+x}\right)}{x}dx-\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{2}{1+x}\right)}{1+x}dx$$

$$X=\int_0^1 \frac{\ln(1-x)\ln 2 -\ln(1-x)\ln(1+x)-\ln(1+x)\ln 2+\ln^2(1+x)}{x}dx$$ $$Y=\int_0^1 \frac{\ln(1-x)\ln 2 -\ln(1-x)\ln(1+x)-\ln(1+x)\ln 2+\ln^2(1+x)}{1+x}dx$$

$$I_1=\ln 2\int_0^1 \frac{\ln(1-x)}{x}dx=\color{blue}{-\ln 2 \zeta(2)}$$ $$I_2=-\int_0^1 \frac{\ln(1-x)\ln(1+x)}{x}dx=\color{red}{\frac{5}{8}\zeta(3)}$$ $$I_3=-\ln 2 \int_0^1 \frac{\ln(1+x)}{x}dx=\color{blue}{-\frac{\ln 2}{2}\zeta(2)}$$ $$I_4=\int_0^1 \frac{\ln^2(1+x)}{x}dx=\color{red}{\frac{\zeta(3)}{4}}$$ $$I_5=\ln 2\int_0^1 \frac{\ln(1-x)}{1+x}dx=\cancel{\frac{\ln^3 2}{2}}-\cancel{\ln 2\frac{\zeta(2)}{2}}$$ $$I_6=-\int_0^1 \frac{\ln(1-x)\ln(1+x)}{1+x}dx =\cancel{-\frac{\ln^3 2}{3}}+\cancel{\ln 2\frac{\zeta(2)}{2}}-\color{red}{\frac{\zeta(3)}{8}}$$ $$I_7=-\ln 2 \int_0^1 \frac{\ln(1+x)}{1+x}dx=\cancel{-\frac{\ln^3 2}{2}}$$ $$I_8=\int_0^1 \frac{\ln^2(1+x)}{1+x}dx=\cancel{\frac{\ln^3 2}{3}}$$

$$I=X-Y=(I_1+I_2+I_3+I_4)-(I_5+I_6+I_7+I_8)=\boxed{\zeta(3)-\frac32 \ln 2 \zeta(2)}$$

You might need the first generalization from the preprint A simple strategy of calculating two alternating harmonic series generalizations by Cornel Ioan Valean. All the calculations are accomplished by avoiding the evaluation of Euler sums.

$$ \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(m)}}{n}=\frac{(-1)^m}{(m-1)!}\int_0^1\frac{\displaystyle \log^{m-1}(x)\log\left(\frac{1+x}{2}\right)}{1-x}\textrm{d}x $$ $$=\frac{1}{2}\biggr(m\zeta (m+1)-2\log (2) \left(1-2^{1-m}\right) \zeta (m)$$ $$-\sum_{k=1}^{m-2} \left(1-2^{-k}\right)\left(1-2^{1+k-m}\right)\zeta (k+1)\zeta (m-k)\biggr).$$

Consider the integral $$K=\int_0^1\frac{\operatorname{Li}_2(x)}{1+x}\ dx$$

By applying IBP we have

$$K=\ln(2)\zeta(2)+\int_0^1\frac{\ln(1-x)\ln(1+x)}{x}\ dx\tag{1}$$

.

On the other hand

\begin{align} K&=\int_0^1\frac{\operatorname{Li}_2(x)}{1+x}\ dx=\int_0^1\frac{1}{1+x}\left(\int_0^1-\frac{x\ln u}{1-xu}\ du\right)\ dx\\ &=\int_0^1\ln u\left(\int_0^1\frac{-x}{(1+x)(1-ux)}\ dx\right)\ du\\ &=\int_0^1\ln u\left(\frac{\ln2}{1+u}+\frac{\ln(1-u)}{u}-\frac{\ln(1-u)}{1+u}\right)\ du\\ &=\ln2\int_0^1\frac{\ln u}{1+u}\ du+\int_0^1\frac{\ln u\ln(1-u)}{u}\ du-\color{blue}{\int_0^1\frac{\ln u\ln(1-u)}{1+u}\ du}\\ &\overset{\color{blue}{IBP}}{=}\ln2\left(-\frac12\zeta(2)\right)+\zeta(3)\color{blue}{-\int_0^1\frac{\ln u\ln(1+u)}{1-u}du+\int_0^1\frac{\ln(1-u)\ln(1+u)}{u}du}\tag{2} \end{align}

Combining (1) and (2) we get

$$\int_0^1\frac{\ln x\ln(1+x)}{1-x}\ dx=\zeta(3)-\frac32\ln2\zeta(2)$$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x} \over 1-x}\,\dd x = \zeta\pars{3} - {3 \over 2}\,\ln\pars{2}\zeta\pars{2}}:\ \approx 0.5082\ {\Large ?}}$.

In (\ref{1}) and (\ref{2}) I used $\ds{\mrm{Li}_{2}\pars{1} = \zeta\pars{2} = \pi^{2}/6}$ and the known values of $\ds{\mrm{Li}_{2}\pars{1/2}}$ and $\ds{\mrm{Li}_{3}\pars{1/2}}$.