A challenging logarithmic integral $\int_0^1 \frac{\log(1+x)\log(1-x)}{1+x}dx$

Question

While playing around with Mathematica, I found that

$$\int_0^1 \frac{\log(1+x)\log(1-x)}{1+x}dx = \frac{1}{3}\log^3(2)-\frac{\pi^2}{12}\log(2)+\frac{\zeta(3)}{8}$$

Please help me prove this result.

Answer 1

Use your favorite program to compute the indefinite integral in terms of polylogarithms $$\int\frac{\ln(1+x)\ln(1-x)\,dx}{1+x}=\frac{\ln2}{2}\ln^2(1+x)-\ln(1+x)\,\mathrm{Li}_2\left(\frac{1+x}{2}\right)+\mathrm{Li}_3\left(\frac{1+x}{2}\right).$$ [This can be verified by straightforward differentiation].

To compute the definite integral, it suffices to know $\mathrm{Li}_{2,3}\left(\frac12\right)$ and $\mathrm{Li}_{2,3}(1)$. However, the definition of polylogarithm immediately implies $\mathrm{Li}_s(1)=\zeta(s)$. Also, the values $\mathrm{Li}_{2,3}\left(\frac12\right)$ can be found here (formulas (16), (17)).

Answer 2

Substitute $x= \frac{1-t}{1+t}$ to derive

\begin{align} & \int_0^1 \frac{\ln(1+x)\ln(1-x)}{1+x}dx =\int_0^1 \frac{\ln\frac{2t}{1+t}\ln\frac2{1+t}}{1+t}dt\\ =& \int_0^1 \frac{[\ln2-\ln(1+t)]^2}{1+t}dt +\ln2 \int_0^1 \frac{\ln t}{1+t}dt -\int_0^1 \frac{\ln t\ln(1+t)}{1+t}dt\\ =&\frac13\ln^32-\frac{\pi^2}{12}\ln2+\frac18\zeta(3) \end{align} where $\int_0^1 \frac{\ln t}{1+t}dt=-\frac{\pi^2}{12}$ and $\int_0^1 \frac{\ln^2(1+t)}{t}dx={\frac{\zeta(3)}{4}}$

$$\int_0^1 \frac{\ln t\ln(1+t)}{1+t}dt\overset{IBP}= -\frac12\int_0^1 \frac{\ln^2(1+t)}tdt = -\frac18\zeta(3) $$

Answer 3

It is easy to prove that $$\int\frac{\log(1-x)}{x+1}dx=\log(2)\log(x+1)-\operatorname{Li}_2\left(\frac{x+1}{2}\right)$$

The proof is straightforward if we set $1+x=y$ and rest it follows by definition of dilogarithm and logarithmic integrals. Next we do integration by parts giving us $$\log(1+x)\left(\log(2)\log(1+x)-\operatorname{Li}_2\left(\frac{x+1}{2}\right)\right)\Bigg|_0^1-\int_0^1\frac{dx}{1+x}\left(\log(2)\log(1+x)-\operatorname{Li}_2\left(\frac{x+1}{2}\right)\right)=\log^3(2)-\frac{\pi^2}{6}\log(2)-\frac{\log^3(2)}{3}+\int_0^1\frac{\operatorname{Li}_2\left(\frac{x+1}{2}\right)}{x+1}dx=\frac{\log^3(2)}{2}-\frac{\pi^2}{6}\log(2)+\zeta(3)-\hbox{Li}_3\left(\frac{1}{2}\right)$$ Now using the known value of $\hbox{Li}_3\left(\frac{1}{2}\right)=\frac{\log^3(2)}{6}-\frac{\pi^2}{12}\log(2)+\frac{7}{8}\zeta(3)$ and simplifying gives us the desires closed form for it. $$\frac{\log^3(2)}{3} -\frac{\pi^2}{12}\log(2)+\frac{\zeta(3)}{8}$$.