Closed form of $\int_0^1(\ln(1-x)\ln(1+x)\ln(x))^2\,dx$

Question

I remember that some time ago I was asking this question Evaluate $\int_0^1\ln(1-x)\ln x\ln(1+x) \mathrm{dx}$ ,
and now, while I was making a review, I asked myself if we can get the closed form of

$$ \int_{0}^{1}\left[\vphantom{\Large A}\ln\left(1 - x\right)\ln\left(1 + x\right)\ln\left(x\right)\right]^{\,2}\,\mathrm{d}x $$

by using the similar tools as in that proof. The problem is that we have to cope with some
series that seem far more complicated. In case you have some fruitful ideas here ...

Answer 1

Process 1:

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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.

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Process 2

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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}

Answer 2

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.

Answer 3

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.