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

Mathematics Asked on November 9, 2021

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.


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. For example: here, here, here, here, here, here, here, here, here, here, here, here, here, here, here, here and here by directly typing them into the program. Also here, here, here, here, here, here, here, here, here after some elementary manipulations (e.g. tangent-half substitution); this and this after $xmapsto 2x/(1+x^2)$ .

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


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.

2 Answers

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}_4left(dfrac12right) & dfrac{pi^4}{90} \[4pt] k=5 & -dfrac{15}{16} zeta(5) & text{Li}_5left(dfrac12right) & 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.

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

Antiderivative 1.0

$$begin{align} &intdfrac{log^3(1+x)}{1-x},text dx = -6text{ Li}_4dfrac {1+x}2 +6text{ Li}_3dfrac{1+x}2log(1+x)\[4pt] &-3text{Li}_2dfrac{1+x}2log^2(1+x) - logdfrac{1-x}2log^3(1+x)+text{const}, end{align}tag3$$

Antiderivative 2.0

$$begin{align} &intdfrac{log^2 (1+x) log(1-x)}{1+x},text dx = -2text{ Li}_4dfrac{1+x} 2 +2text{ Li}_3 dfrac {1+x} 2 log(1+x)\[4pt] &-text{Li}_2 dfrac{1+x} 2log^2(1+x)+dfrac13log(2)log^3(1+x)+text{const}. end{align}tag4$$

Antiderivative 3.0

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)+2text{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)+2text{Li}_3(x)log(x) -text{Li}_2(x)log^2(x)right) frac{log^3(1+x)}{1-x},text dx\[4pt] &-3int_0^1 left(-2,text{Li}_4(x)+2text{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) + 2text{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.

Answered by Yuri Negometyanov on November 9, 2021

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 $$ mathcal{M}[log^n(1pm e^{-x})](s) = (-1)^n n! Gamma(s) S_{s,n}(mp 1) $$ 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 $$ log^n(1pm e^{-x}) = sum_{k=0}^infty frac{(-1)^{k+n} n!}{k!} S_{-k,n}(mp 1)x^k $$ 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 $$ -8cdot3 cdot 19 S_{2,2}(1)S_{1,3}(-1) = -frac{19}{15} pi ^4 text{Li}_4left(frac{1}{2}right)-frac{133}{120} pi ^4 zeta (3) log (2)+frac{19 pi ^8}{1350}+frac{19}{360} pi ^6 log ^2(2)-frac{19}{360} pi ^4 log ^4(2) $$ 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?

Answered by Benedict W. J. Irwin on November 9, 2021

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