Closed form of $\int_0^1(\ln(1-x)\ln(1+x)\ln(x))^2\,dx$
Process 1:
It is known that \begin{align} \int_{0}^{1} x^{\nu -1} (1+x)^{\lambda} (1-x)^{\mu -1} \, dx = B(\mu, \nu) \, {}_{2}F_{1}(- \lambda, \nu; \mu+\nu; -1) \end{align} for which \begin{align} \int_{0}^{1} \left[ \ln(x) \, \ln(1-x) \, \ln(1+x) \right]^{2} \, dx = \partial_{\nu}^{2} \partial_{\mu}^{2} \partial_{\lambda}^{2} \left[ B(\mu, \nu) \, {}_{2}F_{1}(- \lambda, \nu; \mu+\nu; -1) \right]_{\mu=\nu=1}^{\lambda = 0} \end{align} The resulting value will then lead to the calculation of series involving the polygamma function of up to order three.
Process 2
In the view of using the series \begin{align} \frac{1}{2} \, \ln^{2}(1-x) = \sum_{n=1}^{\infty} \frac{H_{n} \, x^{n+1}}{n+1} \end{align} then \begin{align} \frac{1}{4} \, \ln^{2}(1-x) \, \ln^{2}(1+x) = \sum_{n=1}^{\infty} A_{n} \, x^{n+2} \end{align} where \begin{align} A_{n} = \sum_{s=1}^{n} \frac{(-1)^{s-1} \, H_{n-s} H_{s} }{(s+1) (n-s+1)}. \end{align} Now consider the integral \begin{align} J = \int_{0}^{1} x^{\mu + \nu} \, dx = \frac{1}{\mu+\nu+1} \end{align} for which \begin{align} \partial_{\mu}^{2} J = \int_{0}^{1} x^{\mu+\nu} \, \ln^{2}(x) \, dx = \frac{2}{(\mu+\nu+1)^{3}}. \end{align} Now, for the integral \begin{align} I = \int_{0}^{1} \left[ \ln(1-x) \, \ln(1+x) \, \ln(x) \right]^{2} \, dx, \end{align} it is seen that \begin{align} I &= 4 \sum_{n=1}^{\infty} A_{n} \, \int_{0}^{1} x^{n+2} \, \ln^{2}(x) \, dx \\ &= 8 \, \sum_{n=1}^{\infty} \frac{A_{n}}{(n+3)^{3}} \\ &= 8 \, \sum_{n=1}^{\infty} \sum_{s=1}^{\infty} \frac{(-1)^{s-1} \, H_{s} H_{n}}{(n+1) (s+1) (n+s+3)^{3}}. \end{align}
In terms of multiple zeta values this integral is equal to $$ 2 \zeta (\bar3,\bar1)+\zeta (\bar3,1)+6 \zeta (\bar2,\bar1)+3 \zeta (\bar2,1)+24 \zeta (\bar1,\bar1)+12 \zeta (\bar1,1)-6 \zeta (2,\bar1)-2 \zeta (3,\bar1)+\zeta(\bar3,\bar1,\bar1)+\zeta (\bar3,\bar1,1)+\zeta (\bar3,1,\bar1)+2 \zeta (\bar2,\bar1,\bar1)+2 \zeta (\bar2,\bar1,1)+2 \zeta (\bar2,1,\bar1)+6 \zeta (\bar1,\bar1,\bar1)+6 \zeta(\bar1,\bar1,1)+6 \zeta (\bar1,1,\bar1)-2 \zeta (2,\bar1,\bar1)-2 \zeta (2,\bar1,1)-2 \zeta (2,1,\bar1)-\zeta (3,\bar1,\bar1)-\zeta (3,\bar1,1)-\zeta(3,1,\bar1)+\zeta (\bar3,\bar1,\bar1,\bar1)+\zeta (\bar3,\bar1,1,\bar1)+\zeta (\bar3,1,\bar1,1)+\zeta (\bar2,\bar1,\bar1,\bar1)+\zeta (\bar2,\bar1,1,\bar1)+\zeta (\bar2,1,\bar1,1)+2\zeta (\bar1,\bar1,\bar1,\bar1)+2 \zeta (\bar1,\bar1,1,\bar1)+2 \zeta (\bar1,1,\bar1,1)-\zeta (2,\bar1,\bar1,\bar1)-\zeta (2,\bar1,1,\bar1)-\zeta (2,1,\bar1,1)-\zeta(3,\bar1,\bar1,\bar1)-\zeta (3,\bar1,1,\bar1)-\zeta (3,1,\bar1,1)-6 \zeta (3)+3 \zeta (\bar3)+12 \zeta (\bar2)+60 \zeta (\bar1)-12 \zeta (2)-\frac{1}{4}\zeta (4)+90 $$ Some of these may not have a closed form.
Explanation. Every logarithm can be written as a single integral: $$ \log x = -\int_x^1 \frac{dt}{t}, \qquad \log(1+a x) = \int_0^x \frac{du}{u+a^{-1}}. $$ In turn, every iterated integral over a simplex $0<t_n<\ldots<t_1<1$ of the form $$ \int_0^1\frac{dt_1}{t_1-b_1}\int_0^{t_2}\frac{dt_2}{t_2-b_2}\int\cdots\int_0^{t_{n-1}}\frac{dt_n}{t_n-b_n} $$ can be written as a multiple zeta value when each $b$ is in $\{0,\pm1\}$.
The integral has the form $$ \int_0^1dx\int_x^1\frac{dt_1}{t_1}\int_x^1 \frac{dt_2}{t_2} \int_0^x \frac{du_1}{u_1-1} \int_0^x \frac{du_2}{u_2-1} \int_0^x \frac{dv_1}{v_1+1} \int_0^x \frac{dv_2}{v_2+1}, $$ which can be brought into the interated-integral form above by splitting the integration domain into simplices and repeatedly integrating over some variables until only integrands of the form $\frac{dt}{t-b}$ remain. Carrying out this procedure gives the expression above in terms of multiple zeta values.
See this article. The integral is recorded as the most complicated result:
- $\scriptsize \int_0^1 \log ^2(1-x) \log ^2(x) \log ^2(x+1) \, dx=4 \zeta(\bar5,1)-\frac{16}{3} \pi ^2 \text{Li}_4\left(\frac{1}{2}\right)-32 \text{Li}_4\left(\frac{1}{2}\right)+24 \text{Li}_5\left(\frac{1}{2}\right)-16 \text{Li}_6\left(\frac{1}{2}\right)-8 \text{Li}_4\left(\frac{1}{2}\right) \log ^2(2)+24 \text{Li}_4\left(\frac{1}{2}\right) \log (2)-16 \text{Li}_5\left(\frac{1}{2}\right) \log (2)+\frac{275 \zeta (3)^2}{8}+\frac{34 \pi ^2 \zeta (3)}{3}-237 \zeta (3)-\frac{341 \zeta (5)}{2}+7 \zeta (3) \log ^3(2)-\frac{91}{2} \zeta (3) \log ^2(2)-\frac{77}{6} \pi ^2 \zeta (3) \log (2)+156 \zeta (3) \log (2)+\frac{217}{2} \zeta (5) \log (2)+\frac{101 \pi ^6}{5040}-40 \pi ^2-\frac{29 \pi ^4}{45}+720-\frac{2}{9} \log ^6(2)+\frac{4 \log ^5(2)}{5}-\frac{1}{18} \pi ^2 \log ^4(2)+\frac{2 \log ^4(2)}{3}+\frac{2}{3} \pi ^2 \log ^3(2)-24 \log ^3(2)-\frac{1}{36} \pi ^4 \log ^2(2)-8 \pi ^2 \log ^2(2)+144 \log ^2(2)+\frac{2}{3} \pi ^4 \log (2)+32 \pi ^2 \log (2)-480 \log (2)$
A bonus:
- $\scriptsize \int_0^1 \frac{\log ^2(1-x) \log ^2(x) \log ^2(x+1)}{(1-x)^2 x^2 (x+1)^2} \, dx=8 \zeta(\bar5,1)+15\zeta(\bar5,1,1)-\frac{15}{2}\zeta(5,\bar1,1)+\frac{15}{2} \log (2) \zeta(\bar5,1)+15 \text{Li}_4\left(\frac{1}{2}\right) \zeta (3)-\frac{17}{12} \pi ^2 \text{Li}_4\left(\frac{1}{2}\right)+14 \text{Li}_4\left(\frac{1}{2}\right)+\frac{9}{2} \pi ^2 \text{Li}_5\left(\frac{1}{2}\right)+30 \text{Li}_5\left(\frac{1}{2}\right)+16 \text{Li}_6\left(\frac{1}{2}\right)+8 \text{Li}_4\left(\frac{1}{2}\right) \log ^2(2)+30 \text{Li}_4\left(\frac{1}{2}\right) \log (2)+16 \text{Li}_5\left(\frac{1}{2}\right) \log (2)+\frac{239 \pi ^2 \zeta (3)}{48}-\frac{5807 \pi ^2 \zeta (5)}{256}+\frac{9087 \zeta (7)}{32}-\frac{1457 \zeta (5)}{16}-\frac{5417 \zeta (3)^2}{128}-\frac{1259 \pi ^4 \zeta (3)}{1280}+\frac{13}{4} \zeta (3) \log ^4(2)-\frac{49}{4} \zeta (3) \log ^3(2)-\frac{61}{8} \pi ^2 \zeta (3) \log ^2(2)-\frac{203}{8} \zeta (3) \log ^2(2)+\frac{4743}{64} \zeta (5) \log ^2(2)+\frac{123}{2} \zeta (3)^2 \log (2)+\frac{31}{3} \pi ^2 \zeta (3) \log (2)-\frac{57}{4} \zeta (3) \log (2)-\frac{4867}{32} \zeta (5) \log (2)+\frac{167 \pi ^6}{2520}+\frac{19 \pi ^4}{1440}+\frac{2 \log ^6(2)}{9}-\frac{3}{80} \pi ^2 \log ^5(2)+\log ^5(2)-\frac{29}{288} \pi ^2 \log ^4(2)-\frac{17 \log ^4(2)}{12}-\frac{1}{32} \pi ^4 \log ^3(2)+\frac{5}{6} \pi ^2 \log ^3(2)+\frac{235}{576} \pi ^4 \log ^2(2)+\frac{5}{4} \pi ^2 \log ^2(2)+\frac{13}{48} \pi ^4 \log (2)-\frac{521 \pi ^6 \log (2)}{13440}$
This is not easy via elemenatry means. For reference, see here.