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This is a follow-up question to the following questions:

Evaluating $\int_0^1 \frac{\ln^m (1+x)\ln^n x}{x}\; dx$ for $m,n\in\mathbb{N}$

Closed form for ${\large\int}_0^1\frac{\ln^4(1+x)\ln x}x \, dx$

What is a closed form for ${\large\int}_0^1\frac{\ln^3(1+x)\,\ln^2x}xdx$?

Let $p\ge 1$ and $q\ge 1$ be integers. We consider the following quantity: \begin{equation} {\mathfrak I}^{(p,q)}:= \int\limits_0^1 \frac{[\log(1+x)]^p}{x} [\log(x)]^q dx \end{equation}

By using the techniques developed in the questions above we computed the result for $p+q=5$. We have: \begin{eqnarray} {\mathfrak I}^{(5,0)} &=& -120 \text{Li}_6\left(\frac{1}{2}\right)-60 \text{Li}_4\left(\frac{1}{2}\right) \log ^2(2)-120 \text{Li}_5\left(\frac{1}{2}\right) \log (2)-\frac{35}{2} \zeta (3) \log ^3(2)+\frac{8 \pi ^6}{63}-\frac{5 \log ^6(2)}{3}+\frac{5}{4} \pi ^2 \log ^4(2)\\ {\mathfrak I}^{(4,1)} &=& -120 \text{Li}_6\left(\frac{1}{2}\right)-24 \text{Li}_4\left(\frac{1}{2}\right) \log ^2(2)-72 \text{Li}_5\left(\frac{1}{2}\right) \log (2)+12 \zeta (3)^2-3 \zeta (3) \log ^3(2)-2 \pi ^2 \zeta (3) \log (2)+\frac{3}{4} \zeta (5) \log (2)+\frac{26 \pi ^6}{315}-\frac{17 \log ^6(2)}{30}+\frac{1}{3} \pi ^2 \log ^4(2)-\frac{1}{60} \pi ^4 \log ^2(2)+24 {\bf H}^{(1)}_5(1/2) \\ {\mathfrak I}^{(3,2)} &=& -108 \text{Li}_6\left(\frac{1}{2}\right)-36 \text{Li}_5\left(\frac{1}{2}\right) \log (2)+12 \zeta (3)^2+6 \zeta (3) \log ^3(2)-3 \pi ^2 \zeta (3) \log (2)+\frac{9}{8} \zeta (5) \log (2)+\frac{143 \pi ^6}{2520}+\frac{3 \log ^6(2)}{20}-\frac{1}{4} \pi ^2 \log ^4(2)-\frac{1}{40} \pi ^4 \log ^2(2)+36 {\bf H}^{(1)}_5(1/2)\\ {\mathfrak I}^{(2,3)} &=& -72 \text{Li}_6\left(\frac{1}{2}\right)-8 \text{Li}_5\left(\frac{1}{2}\right) \log (8)+6 \zeta (3)^2+4 \zeta (3) \log ^3(2)-\pi ^2 \zeta (3) \log (4)+\frac{3}{4} \zeta (5) \log (2)+\frac{17 \pi ^6}{420}+\frac{\log ^6(2)}{10}-\frac{1}{6} \pi ^2 \log ^4(2)-\frac{1}{60} \pi ^4 \log ^2(2)+24 {\bf H}^{(1)}_5(1/2)\\ {\mathfrak I}^{(1,4)} &=& \frac{93}{4} \text{Li}_6\left(1\right) \end{eqnarray}

Here \begin{equation} {\bf H}^{(p)}_q(x) := \sum\limits_{m=1}^\infty \frac{H_m^{(p)}}{m^q} x^m \end{equation}

Note that the term ${\bf H}^{(1)}_5(1/2)$ cannot be reduced to poly-logarithms only for the following reason. Clearly we have: \begin{eqnarray} &&{\bf H}^{(1)}_5(-1) = \int\limits_0^{-1} \frac{\log((-1)/t)^4}{(4)!}\cdot \frac{Li_1(t)}{t(1-t)} dt\\ &&\underbrace{=}_{u=\frac{t}{t-1}} \frac{1}{4!}\sum\limits_{p=0}^4 \binom{4}{p} (-1)^p \int\limits_0^{1/2} \frac{\log(u)^p \log(1-u)^{5-p}}{u} du\\&&= -2 {\bf H}^{(1)}_5(1/2)+6 \text{Li}_6\left(\frac{1}{2}\right)-2 \text{Li}_4\left(\frac{1}{2}\right) \log ^2(2)+\text{Li}_4\left(\frac{1}{2}\right) \log (2) \log (4)+\text{Li}_5\left(\frac{1}{2}\right) \log (4)-\frac{\zeta (3)^2}{2}+\frac{1}{72} \pi ^2 \left(12 \zeta (3) \log (2)+\log ^4(2)\right)-\frac{1}{3} \zeta (3) \log ^3(2)-\frac{1}{16} \zeta (5) \log (2)-\frac{19 \pi ^6}{4320}-\frac{\log ^6(2)}{120}+\frac{1}{720} \pi ^4 \log ^2(2) \end{eqnarray} Since ${\bf H}^{(1)}_5(-1) = \zeta(-5,1)+Li_6(-1)$ and since it is known that $\zeta(-5,1)$ cannot be reduced to univariate zeta functions the same holds for ${\bf H}^{(1)}_5(1/2)$. To reiterate the quantity ${\bf H}^{(1)}_5(1/2)$ is not redundant in here.

Now my question would be the usual one, meaning can we derive a closed-form expression for the quantity above for arbitary values of $p$ and $q$. From the results above we can see that some new quantity ${\bf H}^{(1)}_5(1/2)$ enters the result. Can this quantity be reduced to polylogarithms and elementary functions? If not then, for $p+q \ge 5$, what will be the minimal set of quantities that will appear in the result?

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  • $\begingroup$ You may like to see arxiv 1910.12113 for higher weights. $\endgroup$ Apr 27, 2020 at 12:42

2 Answers 2

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \left.\vphantom{\Large A}\mc{I}^{\pars{p,q}}\right\vert_{\ p,\, q\ \in\ \mathbb{N}_{\ \geq 1}} & \equiv \int_{0}^{1}{\ln^{p}\pars{1 + x} \over x}\,\ln^{q}\pars{x}\,\dd x \,\,\,\stackrel{x + 1\ \mapsto\ x}{=}\,\,\, \int_{1}^{2}{\ln^{p}\pars{x}\ln^{q}\pars{x - 1} \over x - 1}\,\dd x \\[5mm] &\stackrel{x\ \mapsto\ 1/x}{=}\,\,\, \int_{1}^{1/2}{\ln^{p}\pars{1/x}\ln^{q}\pars{1/x - 1} \over 1/x - 1}\, \pars{-\,{\dd x \over x^{2}}} \\[5mm] & = \pars{-1}^{\,p}\int_{1/2}^{1}{\ln^{p}\pars{x} \bracks{\ln\pars{1 - x} - \ln\pars{x}}^{\,q} \over x\pars{1 - x}}\,\dd x \\[5mm] & = \pars{-1}^{\,p}\sum_{k = 0}^{q}{q \choose k}\pars{-1}^{k} \int_{1/2}^{1}{\ln^{p + k}\pars{x}\ln^{q - k}\pars{1 - x} \over x\pars{1 - x}}\,\dd x \\[1cm] & = \pars{-1}^{\,p}\sum_{k = 0}^{q}{q \choose k}\pars{-1}^{k}\,\,\times \\[2mm] & \bracks{% \int_{1/2}^{1}{\ln^{p + k}\pars{x}\ln^{q - k}\pars{1 - x} \over x}\,\dd x + \int_{1/2}^{1}{\ln^{p + k}\pars{x}\ln^{q - k}\pars{1 - x} \over 1 - x}\,\dd x} \label{1}\tag{1} \end{align}

An example !!!:

With \eqref{1} result: \begin{align} \mc{I}^{\pars{5,0}} & = \pars{-1}^{5}\bracks{% \int_{1/2}^{1}{\ln^{5}\pars{x} \over x}\,\dd x + \int_{1/2}^{1}{\ln^{5}\pars{x} \over 1 - x}\,\dd x} \\[1cm] & = -\left(\vphantom{\huge A}-\,{1 \over 6}\,\ln^{6}\pars{1 \over 2}\right. \\[2mm] & \left.\phantom{-\left(AA\right.} + \braces{\ln\pars{1 - {1 \over 2}}\ln^{5}\pars{1 \over 2} + \int_{1/2}^{1}\ln\pars{1 - x}\bracks{5\ln^{4}\pars{x}\,{1 \over x}}\,\dd x} \right) \\[1cm] & = -\,{5 \over 6}\,\ln^{6}\pars{2} + 5\int_{1/2}^{1}\mrm{Li}_{2}'\pars{x}\ln^{4}\pars{x}\,\dd x \\[5mm] & = -\,{5 \over 6}\,\ln^{6}\pars{2} - 5\,\mrm{Li}_{2}\pars{1 \over 2}\ln^{4}\pars{1 \over 2} - 20\int_{1/2}^{1}\mrm{Li}_{3}'\pars{x}\ln^{3}\pars{x}\,\dd x \\[5mm] & = -\,{5 \over 6}\,\ln^{6}\pars{2} - 5\ln^{4}\pars{2}\,\mrm{Li}_{2}\pars{1 \over 2} + 20\,\mrm{Li}_{3}\pars{1 \over 2}\ln^{3}\pars{1 \over 2} + 60\int_{1/2}^{1}\mrm{Li}_{4}'\pars{x}\ln^{2}\pars{x}\,\dd x \\[1cm] & = -\,{5 \over 6}\,\ln^{6}\pars{2} - 5\ln^{4}\pars{2}\,\mrm{Li}_{2}\pars{1 \over 2} - 20\ln^{3}\pars{2}\,\mrm{Li}_{3}\pars{1 \over 2} - 60\,\mrm{Li}_{4}\pars{1 \over 2}\ln^{2}\pars{1 \over 2} \\[2mm] & - 120\int_{1/2}^{1}\mrm{Li}_{5}'\pars{x}\ln\pars{x}\,\dd x \\[1cm] & = -\,{5 \over 6}\,\ln^{6}\pars{2} - 5\ln^{4}\pars{2}\,\mrm{Li}_{2}\pars{1 \over 2} - 20\ln^{3}\pars{2}\,\mrm{Li}_{3}\pars{1 \over 2} - 60\ln^{2}\pars{2}\,\mrm{Li}_{4}\pars{1 \over 2} \\[2mm] & + 120\,\mrm{Li}_{5}\pars{1 \over 2}\ln\pars{1 \over 2} + 120\int_{1/2}^{1}\mrm{Li}_{6}'\pars{x}\,\dd x \\[1cm] & = -\,{5 \over 6}\,\ln^{6}\pars{2} - 5\ln^{4}\pars{2}\,\mrm{Li}_{2}\pars{1 \over 2} - 20\ln^{3}\pars{2}\,\mrm{Li}_{3}\pars{1 \over 2} - 60\ln^{2}\pars{2}\,\mrm{Li}_{4}\pars{1 \over 2} \\[2mm] & - 120\ln\pars{2}\,\mrm{Li}_{5}\pars{1 \over 2} + 120\,\ \underbrace{\zeta\pars{6}}_{\ds{\pi^{6} \over 945}}\ -\ 120\,\mrm{Li}_{6}\pars{1 \over 2} \end{align}

Since $\ds{\mrm{Li}_{2}\pars{1 \over 2} = {\pi^{2} \over 12} - {\ln^{2}\pars{2} \over 2}}$ and $\ds{\mrm{Li}_{3}\pars{1 \over 2} = {\ln^{3}\pars{2} \over 6} - {\pi^{2}\ln\pars{2} \over 12} + {7\,\zeta\pars{3} \over 8}}$:

$$ \begin{array}{|rcl|}\hline \mbox{}\\ \ds{\mc{I}^{\pars{5,0}}} & \ds{=} & \ds{\int_{0}^{1}{\ln^{5}\pars{1 + x} \over x}\,\dd x} \\[2mm] & \ds{=} & \ds{{8\pi^{6} \over 63} + {5\pi^{2}\ln^{4}\pars{2} \over 4} - {5\ln^{6}\pars{2} \over 3} - {35\ln^{3}\pars{2}\zeta\pars{3} \over 2}} \\[2mm] && \ds{- 60\ln^{2}\pars{2}\mrm{Li}_{4}\pars{1 \over 2}- 120\ln\pars{2}\mrm{Li}_{5}\pars{1 \over 2}- 120\,\mrm{Li}_{6}\pars{1 \over 2} \approx 0.0422} \\ &&\mbox{}\\ \hline \end{array} $$

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  • $\begingroup$ Yes, thank you for this. I have figured out myself that a substituting for $1/(1+x)$ in the original integral reduces it to integrals $\int\limits_0^{1/2} \log(x)^p \log(1-x)^q dx$ for $p+q=const$ you quoted above. The essential difficulty is to compute the later integrals. $\endgroup$
    – Przemo
    Dec 18, 2017 at 11:24
  • $\begingroup$ @Przemo You're right. The general 'trick' is, somehow, to switch $x + 1$ to $1 - x$ and divide by some power of $x$. Of course, whenever it's possible. Thanks. $\endgroup$ Dec 18, 2017 at 15:34
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It's not an answer but it's too lenghty to be a comment.

$\begin{align}{\mathfrak I}^{(4,1)}&=\frac{35}{6}\operatorname{Li_6}\left(\frac{1}{2}\right)+\frac{1}{4}\operatorname{Li_5}\left(\frac{1}{2}\right)\ln 2-\frac{157}{12}\operatorname{Li_4}\left(\frac{1}{2}\right)\ln^2 2+\frac{43}{12}\zeta(3)^2-\frac{35}{6}\zeta(3)\ln^3 2\\ &-\frac{11}{2}\pi^2\zeta(3)\ln 2-\frac{451}{24}\zeta(5)\ln 2+\frac{1}{8}\pi^6+\frac{1}{3}\ln^6 2+\frac{23}{24}\pi^2\ln^4 2-\frac{17}{12}\pi^4\ln^2 2\end{align}$

I think the extra ${\bf H}^{(p)}_q$ function is probably useless.

${\mathfrak I}^{(p,q)}$ is probably a linear rational combination of,

$\displaystyle \prod_{k=1}^m\zeta(a_k)\ln^r 2\prod_{k=1}^s\operatorname{Li}_{b_k}\left(\frac{1}{2}\right)$

$(a_1+...+a_m)+r+(b1+...+b_s)=p+q+1$

PS: I dont have a proof for the formula above.

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    $\begingroup$ Your formula is not exact but instead it is only some sort of approximation to twenty nine digits decimal digits. One can come up with loads of similar "equalities" just propping up the number of decimal digits in wayback.cecm.sfu.ca/cgi-bin/EZFace/zetaform.cgi .The term ${\bf H}^{(1,5)}(1/2)$ is not redundant and it cannot be reduced to poly-logarithms only as I will explain below. $\endgroup$
    – Przemo
    Dec 18, 2017 at 11:06

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