Evaluate $\int_0^1\ln(1-x)\ln x\ln(1+x)\,\mathrm dx$ What would you recommend me for the integral below?
$$
\int_{0}^{1}\ln(1 - x)\ln(x)
\ln(1 + x)\,\mathrm dx
$$
For instance, for the version without the last logarithm would work to use Taylor series, but in this case things are a bit more complicated and it doesn't seem to work.
 A: \begin{align}
\log(1+x)\log(1-x) & = \left(\sum_{k=1}^{\infty} \dfrac{(-x)^k}k \right)\left(\sum_{k=1}^{\infty} \dfrac{x^k}k \right)\\
& = \sum_{k=1}^{\infty}\sum_{m=1}^{\infty}  \dfrac{(-1)^k x^{k+m}}{km}\\
& = \sum_{l=2}^{\infty} \left(\sum_{k=1}^{l-1} \dfrac{(-1)^k}{k(l-k)} \right) x^l \\
& = \sum_{l=1}^{\infty} \left(\sum_{k=1}^{2l-1} \dfrac{(-1)^k}{k(2l-k)} \right) x^{2l}
\end{align}
Now $$\int_0^1 x^{2l} \log(x) dx = -\dfrac1{(2l+1)^2}$$
Hence, the integral is
$$\sum_{l=2}^{\infty} \left(\sum_{k=1}^{l-1} \dfrac{(-1)^k}{k(l-k)} \right) x^l = \sum_{l=1}^{\infty} \left(\sum_{k=1}^{2l-1} \dfrac{(-1)^{k-1}}{k(2l-k)} \right)\dfrac1{(2l+1)^2}$$
P.S: Bit too long for a comment.
A: Mathematica gives for the final result:
$$\frac{21 \zeta (3)}{8}-6-\log ^2(2)+\log (16)-\frac{1}{12} \pi ^2 (\log(64)-5)$$
The indefinite integral is really really long..
Regarding a series what's wrong with a Maclaurin series around, say, $x=1/e$?
A: $$I=\int _0^1\ln \left(x\right)\ln \left(1-x\right)\ln \left(1+x\right)\:dx$$
By using the the algebraic identity $ab=\frac{1}{4}\left(a+b\right)^2-\frac{1}{4}\left(a-b\right)^2$ we get:
$$I=\frac{1}{4}\underbrace{\int _0^1\ln \left(x\right)\ln ^2\left(1-x^2\right)\:dx}_{t=x^2}-\frac{1}{4}\int _0^1\ln \left(x\right)\ln ^2\left(\frac{1-x}{1+x}\right)\:dx$$
$$=\underbrace{\frac{1}{16}\int _0^1\frac{\ln \left(t\right)\ln ^2\left(1-t\right)}{\sqrt{t}}\:dt}_{I_1}-\underbrace{\frac{1}{4}\int _0^1\ln \left(x\right)\ln ^2\left(1-x\right)\:dx}_{I_2}+\frac{1}{2}\underbrace{\int _0^1\ln \left(x\right)\ln \left(1-x\right)\ln \left(1+x\right)\:dx}_{I}-\underbrace{\frac{1}{4}\int _0^1\ln \left(x\right)\ln ^2\left(1+x\right)\:dx}_{I_3}$$
$$\frac{1}{2}I=\frac{1}{16}I_1-\frac{1}{4}I_2-\frac{1}{4}I_3$$

$$I_1=\frac{1}{16}\int _0^1\frac{\ln \left(t\right)\ln ^2\left(1-t\right)}{\sqrt{t}}\:dt=\frac{1}{16}\lim_{\alpha\rightarrow 1/2\\\beta\rightarrow 1}\frac{\partial ^3}{\partial \alpha \partial \beta ^2}\text{B}\left(\alpha ,\beta \right)$$
$$=\frac{7}{4}\zeta \left(3\right)-6-\ln ^2\left(2\right)+4\ln \left(2\right)+2\zeta \left(2\right)-\frac{3}{2}\ln \left(2\right)\zeta \left(2\right)$$

$$I_2=-\frac{1}{4}\underbrace{\int _0^1\ln \left(x\right)\ln ^2\left(1-x\right)\:dx}_{t=1-x}=-\frac{1}{4}\int _0^1\ln \left(1-t\right)\ln ^2\left(t\right)\:dt=\frac{1}{2}\sum _{k=1}^{\infty }\frac{1}{k\left(k+1\right)^3}$$
$$=-\frac{1}{2}\zeta \left(3\right)+\frac{3}{2}-\frac{1}{2}\zeta \left(2\right)$$

For this one make use of $\ln ^2\left(1-x\right)=2\sum _{k=1}^{\infty }\left(\frac{H_k}{k}-\frac{1}{k^2}\right)x^k$.
$$I_3=-\frac{1}{4}\int _0^1\ln \left(x\right)\ln ^2\left(1+x\right)\:dx$$
$$=-\frac{1}{2}\sum _{k=1}^{\infty }\frac{\left(-1\right)^kH_k}{k}\int _0^1x^k\ln \left(x\right)\:dx+\frac{1}{2}\sum _{k=1}^{\infty }\frac{\left(-1\right)^k}{k^2}\int _0^1x^k\ln \left(x\right)\:dx$$
$$=\frac{1}{2}\sum _{k=1}^{\infty }\frac{\left(-1\right)^kH_k}{k\left(k+1\right)^2}-\frac{1}{2}\sum _{k=1}^{\infty }\frac{\left(-1\right)^k}{k^2\left(k+1\right)^2}$$
$$=\frac{1}{16}\zeta \left(3\right)-\frac{1}{4}\zeta \left(2\right)+\frac{1}{2}\ln ^2\left(2\right)+\frac{3}{2}-2\ln \left(2\right)$$

Collecting the results after multiplying by $2$ we have:
$$I=\int _0^1\ln \left(x\right)\ln \left(1-x\right)\ln \left(1+x\right)\:dx$$
$$=\frac{21}{8}\zeta \left(3\right)-3\ln \left(2\right)\zeta \left(2\right)+\frac{5}{2}\zeta \left(2\right)-\ln ^2\left(2\right)-6+4\ln \left(2\right)$$
