Remarkable logarithmic integral $\int_0^1 \frac{\log^2 (1-x) \log^2 x \log^3(1+x)}{x}dx$

Question

We have the following result ($\text{Li}_{n}$ being the polylogarithm):

$$\tag{*}\small{ \int_0^1 \log^2 (1-x) \log^2 x \log^3(1+x) \frac{dx}{x} = -168 \text{Li}_5(\frac{1}{2}) \zeta (3)+96 \text{Li}_4(\frac{1}{2}){}^2-\frac{19}{15} \pi ^4 \text{Li}_4(\frac{1}{2})+\\ 12 \pi ^2 \text{Li}_6(\frac{1}{2})+8 \text{Li}_4(\frac{1}{2}) \log ^4(2)-2 \pi ^2 \text{Li}_4(\frac{1}{2}) \log ^2(2)+12 \pi ^2 \text{Li}_5(\frac{1}{2}) \log (2)+\frac{87 \pi ^2 \zeta (3)^2}{16}+\\ \frac{447 \zeta (3) \zeta (5)}{16}+\frac{7}{5} \zeta (3) \log ^5(2)-\frac{7}{12} \pi ^2 \zeta (3) \log ^3(2)-\frac{133}{120} \pi ^4 \zeta (3) \log (2)-\frac{\pi ^8}{9600}+\frac{\log ^8(2)}{6}- \\ \frac{1}{6} \pi ^2 \log ^6(2)-\frac{1}{90} \pi ^4 \log ^4(2)+\frac{19}{360} \pi ^6 \log ^2(2) }$$

This is extremely amazing: almost all other similar integrals are not expressible via ordinary polylogarithm.

The solution is however non-trivial. There are two methods: first is to find enough linear relations between similar integrals, once the rank is high enough, solving the system gives $(*)$; second method is to convert the integral into multiple zeta values, then use known linear relations between them. None of these methods can explain the result's simplicity.

Question: Is there a simpler method to prove (*), or a conceptual explanation of its elegance?

Any thought is welcomed. Thank you very much.

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I wrote a Mathematica package, it can calculate the integral in subject and many similar ones. The following command calculates $(*)$:

MZIntegrate[Log[1-x]^2*Log[x]^2*Log[1+x]^3/x, {x,0,1}]

It can also solve some other integrals.

The package can be obtained here. I hope it can benefit those interested in related integral/series.

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Remarks on the question:

• It's known that $\zeta(\bar{3},1,\bar{3},1)$ is very reminiscent to the RHS of $(*)$. But both the simplicity of $\zeta(\bar{3},1,\bar{3},1)$ and its connection to the integral are elusive to me.
• (Added by Iridescent) This contains nearly all known general formulas of these log integrals. However it does not help much on solving OP's problem.

Answer 1

Here are some ideas towards explaining the form of the right hand side. I'm a bit stuck and my main approach hasn't worked out. This may just be rephrasing things in terms of other log-integrals, but hopefully this is a useful way of looking at the problem.

Taking the integral $$ I = \int_0^1 \log^2(1-x) \log^2(x) \log^3(1+x) \frac{dx}{x} $$ we can also rewrite this as $$ I = \int_0^\infty \log^2(1-e^{-x}) \log^2(e^{-x}) \log^3(1+e^{-x}) \; dx $$ which is suited for interpretation as a Mellin transform. Specifically, the power of $x$, is controlled by the power on $\log(x)$ in the original integral format as $$ I = \int_0^\infty x^2 \log^2(1-e^{-x})\log^3(1+e^{-x}) \; dx $$ according to Mathematica we have in general a result for the Mellin transform of the other components $$ \mathcal{M}[\log^n(1\pm e^{-x})](s) = (-1)^n n! \Gamma(s) S_{s,n}(\mp 1) $$ invoking the Neilsen Generalisation of the polylogarithm, $S_{s,n}$. This does recreate the series expansion for $\log(1+e^{-x})$ but the series for $\log(1-e^{-x})$ has a $\log(x)$ term, which might be causing a problem.

We could toy with the idea of a formal series via the Ramanujan Master Theorem, using these Mellin transforms $$ \log^n(1\pm e^{-x}) = \sum_{k=0}^\infty \frac{(-1)^{k+n} n!}{k!} S_{-k,n}(\mp 1)x^k $$ and then the Cauchy product $$ \log^a(1 + e^{-x})\log^b(1 - e^{-x}) = \left( \sum_{k=0}^\infty \frac{(-1)^{k+a} a!}{k!} S_{-k,a}(-1)x^k \right)\left( \sum_{k=0}^\infty \frac{(-1)^{k+b} b!}{k!} S_{-k,b}(1)x^k \right) $$ $$ \log^a(1 + e^{-x})\log^b(1 - e^{-x}) = \sum_{k=0}^\infty \left(\sum_{l=0}^k \frac{(-1)^{a+b+k} a! b!}{l!(k-l)!} S_{-l,a}(-1) S_{l-k,b}(1)\right) x^k $$ alternatively $$ \log^a(1 + e^{-x})\log^b(1 - e^{-x}) = \sum_{k=0}^\infty \frac{(-1)^k}{k!} \left(\sum_{l=0}^k (-1)^{a+b} a! b! \binom{k}{l} S_{-l,a}(-1) S_{l-k,b}(1)\right) x^k $$ plausibly leading to (via RMT) $$ \mathcal{M}\left[ \log^a(1 + e^{-x})\log^b(1 - e^{-x})\right](s) = \Gamma(s) \sum_{l=0}^{-s} (-1)^{a+b} a! b! \binom{-s}{l} S_{-l,a}(-1) S_{l-k,b}(1) $$ then we would conceptually have (with some dodgy negative parts) an answer for the integral as a sum over (four?) pairs of generalized Polylogs, specifically in the case that $s=3$.

This motivates an expression in terms of pairs of $S_{n,k}(z)$, we can guess a term and quickly find $$ -8\cdot3 \cdot 19 S_{2,2}(1)S_{1,3}(-1) = -\frac{19}{15} \pi ^4 \text{Li}_4\left(\frac{1}{2}\right)-\frac{133}{120} \pi ^4 \zeta (3) \log (2)+\frac{19 \pi ^8}{1350}+\frac{19}{360} \pi ^6 \log ^2(2)-\frac{19}{360} \pi ^4 \log ^4(2) $$ this covers a few of the terms in your expression R.H.S. It is likely that other terms contribute to $\pi^8$ for example. I can't get an explicit value for $S_{2,3}(-1)$ to explore this further, but I would assume this holds a $\mathrm{Li}_5(1/2)$ term among others, and the other factor is $S_{1,2}(1) = \zeta(3)$. Perhaps your linear combinations method can be rephrased in terms of the generalized polylogarithm?

Answer 2

Some values of $\text{Li}_k(z)$ are presented in the table $(1).$

\begin{vmatrix} \hspace{-5mu}^{\overline{\hspace{52pt}}}\hspace{-10mu} &\hspace{-10mu}^{\overline{\hspace{64pt}}}\hspace{-10mu} &\hspace{-10mu}^{\overline{\hspace{186pt}}}\hspace{-10mu} &\hspace{-10mu}^{\overline{\hspace{64pt}}}\hspace{-8mu} \\[-4pt] \text{Li}_k(z) & z = -1 & z = \dfrac12 & z = 1 \\[-0pt] \hspace{-5mu}^{\overline{\hspace{52pt}}}\hspace{-10mu} &\hspace{-10mu}^{\overline{\hspace{64pt}}}\hspace{-10mu} &\hspace{-10mu}^{\overline{\hspace{186pt}}}\hspace{-10mu} &\hspace{-10mu}^{\overline{\hspace{64pt}}}\hspace{-8mu} \\[-2pt] k=1 & -\log(2) & \log(2) & \infty \\[4pt] k=2 & -\dfrac{\pi^2}{12} & \dfrac{\pi^2}{12} - \dfrac12 \log^2(2) & \dfrac{\pi^2}{6} \\[4pt] k=3 & -\dfrac34 \zeta(3) & -\dfrac1{12} \pi^2 \log(2) + \dfrac16 \log^3(2) + \dfrac{21}{24} \zeta(3)) & \zeta(3) \\[4pt] k=4 & -\dfrac{7 \pi^4}{720} & \text{Li}_4\left(\dfrac12\right) & \dfrac{\pi^4}{90} \\[4pt] k=5 & -\dfrac{15}{16} \zeta(5) & \text{Li}_5\left(\dfrac12\right) & \zeta(5)\\[-2pt] \hspace{-7mu}\_\_\_\_\_\_\_\_\_\_\_\hspace{-9mu} &\hspace{-9mu}\_\_\_\_\_\_\_\_\_\_\_\_\_\hspace{-9mu} &\hspace{-9mu} \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\hspace{-9mu} & \hspace{-11mu}\_\_\_\_\_\_\_\_\_\_\_\_\_\hspace{-5mu} \tag1 \end{vmatrix}

Also, are known the next antiderivatives below.

$$\int\dfrac{\log (1-x)\log^2 (x)}x\,\text dx = -2\,\text{Li}_4(x)+2\text{Li}_3(x)\log(x)-\text{Li}_2(x)\log^2(x) + \text{const},\tag2$$

$$\begin{align} &\int\dfrac{\log^3(1+x)}{1-x}\,\text dx = -6\text{ Li}_4\dfrac {1+x}2 +6\text{ Li}_3\dfrac{1+x}2\log(1+x)\\[4pt] &-3\text{Li}_2\dfrac{1+x}2\log^2(1+x) - \log\dfrac{1-x}2\log^3(1+x)+\text{const}, \end{align}\tag3$$

$$\begin{align} &\int\dfrac{\log^2 (1+x) \log(1-x)}{1+x}\,\text dx = -2\text{ Li}_4\dfrac{1+x} 2 +2\text{ Li}_3 \dfrac {1+x} 2 \log(1+x)\\[4pt] &-\text{Li}_2 \dfrac{1+x} 2\log^2(1+x)+\dfrac13\log(2)\log^3(1+x)+\text{const}. \end{align}\tag4$$

\begin{align} &\int_0^1 \log(1-x) \log^3(1+x)\,\frac{\log (1-x) \log^2 (x)}{x}\,\text dx \\[5mm] &\overset{IBP(2)}{=\!=\!=\!=}\, \log (1-x) \log^3(1+x) \left(-2\,\text{Li}_4(x)+2\text{Li}_3(x)\log(x)-\text{Li}_2(x)\log^2(x)\right)\bigg|_0^1\\[4pt] &-\int_0^1 \left(-2\,\text{Li}_4(x)+2\text{Li}_3(x)\log(x) -\text{Li}_2(x)\log^2(x)\right) \frac{\log^3(1+x)}{1-x}\,\text dx\\[4pt] &-3\int_0^1 \left(-2\,\text{Li}_4(x)+2\text{Li}_3(1+x)\log(x) -\text{Li}_2(x)\log^2(x)\right) \frac{\log (1- x) \log^2 (1+x)}{1+x} \,\text dx\\[4pt] & \overset{(3),(4)}{=\!=\!=\!=}\, \int_0^1 \left(-2\,\text{Li}_4(x) + 2\text{Li}_3(x)\log(x) - \text{Li}_2(x)\log^2(x)\right)\text{ d}\Bigl(\log(1-x)\log^3(1+x)\Bigr), \end{align} without suitable continuation.