On a certain integral that involves a product of powers of logarithms.

Question

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?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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 $\ds{\mrm{Li}_{3}\pars{1 \over 2} = {\ln^{3}\pars{2} \over 6} - {\pi^{2}\ln\pars{2} \over 12} + {7\,\zeta\pars{3} \over 8}}$:

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

Answer 2

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.