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Can someone compute

$$ \int_0^1\frac{\ln^4(1+x)\ln x}x \,dx$$

in closed form?

I conjecture that the answer can be expressed as a polynomial function with rational coefficients in constants of the form $\operatorname{Li}_n(x)$ where $n$ is a natural number, $x$ is rational, and $\mathrm{Li}_n$ is the $n$th polylogarithm.

The reason for my conjecture is that

$$ \int_0^1\frac{\ln^2(1+x)\ln x}x \; dx =\frac{\pi^4}{24}-\frac16\ln^42+\frac{\pi^2}6\ln^22-\frac72\zeta(3)\ln2-4\operatorname{Li}_4\!\left(\frac12\right) $$

and as shown here on Math StackExchange

$$ \int_0^1\frac{\ln^3(1+x)\ln x}x \; dx = \frac{\pi^2}3\ln^32-\frac25\ln^52+\frac{\pi^2}2\zeta(3)+\frac{99}{16}\zeta(5)-\frac{21}4\zeta(3)\ln^22\\-12\operatorname{Li}_4\left(\frac12\right)\ln2-12\operatorname{Li}_5\left(\frac12\right).$$

The Riemann zeta function obeys

$$ \zeta_n = \operatorname{Li}_n(1) $$

and $\pi^{2n}$ is a rational number times $\zeta_{2n}$. Also,

$$ \operatorname{Li}_1(x) = -\ln(1 - x) .$$

So, the two integrals above are polynomials with rational coefficients in constants of the form $\operatorname{Li}_n(x)$ where $n$ is a natural number, $x$ is rational, and $\mathrm{Li}_n$ is the $n$th polylogarithm. Maybe this pattern continues!

If my conjecture is true, next I'll ask about

$$ \int_0^1\frac{\ln^k(1+x)\ln x}x \,dx$$

for $k = 4, 5, 6, \dots $

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  • $\begingroup$ Your conjecture is based on those two examples? Or do you have more clues to guess the closed form of the integrals you are considering? $\endgroup$ – user37238 Jun 27 '17 at 13:05
  • $\begingroup$ We have $$ \log^4(1-x)=\sum_{n\geq 1}\frac{4x^n}{n}\left[H_{n-1}^3-3 H_{n-1} H_{n-1}^{(2)}+2 H_{n-1}^{(3)}\right] \tag{1}$$ hence the given integral equals $$ I=\sum_{n\geq 1}\frac{4(-1)^{n+1}}{n^3}\left[H_{n-1}^3-3 H_{n-1} H_{n-1}^{(2)}+2 H_{n-1}^{(3)}\right]\tag{2} $$ $\endgroup$ – Jack D'Aurizio Jun 27 '17 at 13:15
  • $\begingroup$ Plenty of Euler sums are involved. What is the purpose of reaching a closed form depending on values of $\zeta$ at the integers and $\text{Li}_n$ at $\frac{1}{2}$? $\endgroup$ – Jack D'Aurizio Jun 27 '17 at 13:16
  • $\begingroup$ Might be a fun problem to apply the PSQL algorithm to: crd-legacy.lbl.gov/~dhbailey/dhbpapers/ppslq.pdf $\endgroup$ – user14717 Jun 27 '17 at 17:24
  • 6
    $\begingroup$ $$ \int_0^1\frac{\ln^4(1+x)\ln x}x \; dx = \frac{\pi^2}3\ln^42-\frac{17}{30}\ln^62-\frac{\pi^4}{60}\ln^22+\frac{26}{315}{\pi^6}-2{\pi^2}\zeta(3)\ln2+12(\zeta(3))^2+\frac{3}{4}\zeta(5)\ln2-3\zeta(3)\ln^32-24\operatorname{Li}_4\left(\frac12\right)\ln^22-72\operatorname{Li}_5\left(\frac12\right)\ln2-96\operatorname{Li}_6\left(\frac12\right)+12S.$$ $$ S=\sum^\infty_{n=1}\frac{H_n}{(n+1)^52^n}=0.017446006115193776854329$$ $\endgroup$ – user178256 Jun 28 '17 at 6:04
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The Stirling numbers of the first kind $\left[ \begin{array}{c} n \\ k \end{array} \right]$ are usually defined by: $$ \sum\limits_{k=0}^n \left[ \begin{array}{c} n \\ k \end{array} \right] x^k := x(x+1)…(x+n-1) $$

The definition of $\enspace\eta_n(m)\enspace$ in $\enspace$ Evaluate $\int_{0}^{\pi }\theta ^{3}\log^{3}\left ( 2\sin\frac{\theta }{2} \right )\mathrm{d}\theta $ is

$$\eta_n(m):=\sum\limits_{k=1}^\infty \frac{(-1)^{k-1}}{k^m}\left(\frac{n!}{(k-1)!}\left[\begin{array}{c} k \\ n+1 \end{array} \right]\right)$$

with $\enspace m>0$, $\enspace n\in\mathbb{N}_0$ , $\enspace\eta_0(m)=\eta(m)$

and $\enspace\displaystyle \frac{1}{(k-1)!} \left[ \begin{array}{c} k \\ {n+1} \end{array} \right]= \sum\limits_{i_1=1}^{k-1}\sum\limits_{i_2=i_1+1}^{k-1}…\sum\limits_{i_n=i_{n-1}+1}^{k-1}\frac{1}{i_1 i_2 … i_n}$ .

To combine this series with the polylogarithm is a separate problem.

Now we can write:

$\displaystyle \sum\limits_{k=1}^\infty \frac{z^k}{k!}\int\limits_0^1 \frac{\ln^k(1+x)\ln x}{x} dx = \int\limits_0^1 \frac{((1+x)^z-1)\ln x}{x} dx = - \sum\limits_{k=1}^\infty \binom z k \frac{1}{k^2}$

$\hspace{5.3cm}\displaystyle = - \sum\limits_{k=1}^\infty z^k \sum\limits_{v=k}^\infty \frac{(-1)^{k-v}}{v^2 v!} \left[ \begin{array}{c} v \\ k \end{array} \right]= \sum\limits_{k=1}^\infty \frac{(-z)^k}{(k-1)!} \eta_{k-1}(3)$

It follows $$\int\limits_0^1 \frac{\ln^k(1+x)\ln x}{x} dx = (-1)^k\,k\,\eta_{k-1}(3)$$ for $k\in\mathbb{N}$ .

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It's not a complete answer but it is too lengthy for a comment.

@user14717: Program GP PARI contains a routine to perform something like PSQL stuff.

Here a script

\p 200

A1=Pi^6

A2=Pi^4*log(2)^2

A3=Pi^2*log(2)^4

A4=polylog(6,1/2)

A5=polylog(5,1/2)*log(2)

A6=polylog(4,1/2)*log(2)^2

A7=polylog(3,1/2)*log(2)^3

A8=polylog(2,1/2)*log(2)^4

A9=log(2)^6

A10=zeta(3)*log(2)^3

A11=zeta(5)*log(2)

A12=zeta(3)^2

A13=Pi^2*log(2)*zeta(3)

A14=polylog(3,1/2)*Pi^2*log(2)

A15=polylog(4,1/2)*Pi^2

A16=polylog(3,1/2)^2

A17=polylog(2,1/2)^2*log(2)^2

A18=polylog(2,1/2)^2*Pi^2

A19=polylog(2,1/2)^3

A20=polylog(2,1/2)*Pi^2*log(2)^2

A21=polylog(2,1/2)*Pi^4

A22=polylog(3,1/2)*zeta(3)

J=intnum(x=0,1,log(1+x)^4*log(x)/x)

lindep([J,A1,A2,A4,A5,A6,A7,A8,A9,A11,A12,A13,A15,A22])

Last command returns an integer relation of Ai's equals to 0. Notice that some Ai's are linearly dependant on integers.

Anyway i wasn't able to find out such integer relation using these constants.

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I am very skeptical as to whether the conjecture is true (except of course when $k=2,3$). Let me explain why. Note that the following identity holds: \begin{eqnarray} &&-2\int\limits_0^1 \frac{\log(1+x)^k}{x} \log(x) dx=\\ &&\imath \pi \sum\limits_{l=1}^{k+1} (-1)^l k_{(l-1)} (Li_l(2)-Li_{k+1}(1)1_{l=k+1}) \log[2]^{k+1-l}+\\ &&\sum\limits_{l=2}^k (-1)^{l-1} k_{(l-1)} \int\limits_1^2 \frac{Li_{l}(x)}{1-x} [\log(x)]^{k+1-l} dx+\\ &&(-1)^{k+1} k! \int\limits_0^1 \frac{Li_{k+1}(1+x)-Li_{k+1}(1)}{x} dx \end{eqnarray} The identity above comes from the knowledge of the anti-derivative of the fraction in the integrand and from integrating by parts once. Now, if we take a look at the last integral above we can actualy write down the antiderivative. We have: \begin{eqnarray} &&\int \frac{Li_{k+1}(1+x)-Li_{k+1}(1)}{x} dx=\\ &&\left\{ \begin{array}{rr} \sum\limits_{l=1}^{k/2} (-1)^l Li_l(1+x) Li_{k+2-l}(1+x) + (-1)^{k/2+1} \frac{1}{2} [Li_{k/2+1}(1+x)]^2 - Li_{k+1}(1) \log(x)& \mbox{if $k$ is even}\\ \sum\limits_{l=1}^{(k+1)/2} (-1)^l Li_l(1+x) Li_{k+2-l}(1+x) + (-1)^{(k+1)/2+1} \int \frac{[Li_{(k+1)/2}(1+x)]^2}{1+x} dx - Li_{k+1}(1) \log(x)& \mbox{if $k$ is odd} \end{array} \right. \end{eqnarray} As we can see it is only when $k$ is even that the result reduces to poly-logarithms only otherwise a new unknown antiderivative involving a square of a polylogarithm remains. As a matter of fact it is not difficult to see that by integrating by parts and changing and swapping order of integration appropriately all the remaining integrals in the third line from the top in the first formula above reduce to poly-logarithms and following quantities: \begin{equation} S^{(2,p)}_q := \int\limits_1^2 \frac{[Li_q(x)]^2}{x}\cdot [\log(x)]^p dx \end{equation} where $p\ge1$ and $q\ge 1$.

Let us now take $k=4$ as an example. Here the integrals in question read: \begin{eqnarray} \int\limits_1^2 \frac{Li_2(x)}{1-x} \cdot [\log(x)]^3 dx &=& -\frac{3}{2} \text{Li}_2(2){}^2 \log ^2(2)-i \pi \text{Li}_2(2) \log ^3(2) + 3 S^{(2,1)}_2 - S^{(2,3)}_1\\ \int\limits_1^2 \frac{Li_3(x)}{1-x} \cdot [\log(x)]^2 dx &=& \text{Li}_3(2){}^2-2 \text{Li}_2(2) \text{Li}_3(2) \log (2)-\zeta (3)^2-\frac{1}{32} \left(\pi ^4+28 i \pi \zeta (3)\right) \log ^2(2) + 3 S^{(2,1)}_2\\ \int\limits_1^2 \frac{Li_4(x)}{1-x} \cdot [\log(x)]^1 dx &=& \text{Li}_3(2){}^2-\frac{\pi ^2 \text{Li}_4(2)}{4}-\text{Li}_2(2) \text{Li}_3(2) \log (2)-\zeta (3)^2+\frac{\pi ^6}{540} + S^{(2,1)}_2\\ \int\limits_0^1 \frac{Li_5(1+x)-Li_5(1)}{x} &=& \frac{1}{2} \left(-\text{Li}_3(2){}^2+2 \text{Li}_2(2) \text{Li}_4(2)+\zeta (3)^2\right)+i \pi (\text{Li}_5(2)-\zeta (5))-\frac{\pi ^6}{540} \end{eqnarray} Now, when we bring everything together the quantities $S^{(2,1)}_2$ miraculously cancel out and then by substituting $x\leftarrow 1/x$ in the quantity $S^{(2,3)}_1$ we obtain the following result: \begin{eqnarray} &&\int\limits_0^1 \frac{[\log(1+x)]^4}{x} \cdot \log(x) dx =\\ &&-\log (2) \left(96 \text{Li}_5\left(\frac{1}{2}\right)+48 \text{Li}_4\left(\frac{1}{2}\right) \log (2)+14 \zeta (3) \log ^2(2)+\log ^5(2)\right)+\\ &&-96 \text{Li}_6\left(\frac{1}{2}\right)+\frac{32 \pi ^6}{315}+\pi ^2 \log ^4(2)+2 \int\limits_{1/2}^1 \frac{[\log(1-x)]^2}{x} \cdot [\log(x)]^3 dx=\\ &&12 \left(\zeta (3)^2-10 \text{Li}_6\left(\frac{1}{2}\right)\right)+\frac{3}{4} \log (2) \left(\zeta (5)-96 \text{Li}_5\left(\frac{1}{2}\right)\right)-24 \text{Li}_4\left(\frac{1}{2}\right) \log ^2(2)+\frac{1}{3} \pi ^2 \left(\log ^4(2)-6 \zeta (3) \log (2)\right)-3 \zeta (3) \log ^3(2)+\frac{26 \pi ^6}{315}-\frac{17 \log ^6(2)}{30}-\frac{1}{60} \pi ^4 \log ^2(2) + 24 {\bf H}^{(1)}_5(1/2) \end{eqnarray} where in the last line we expressed the remaining integral through Euler sums and used the following results: \begin{eqnarray} {\bf H}^{(1)}_2(1/2) &=& \zeta (3)-\frac{1}{12} \pi ^2 \log (2)\\ {\bf H}^{(1)}_3(1/2) &=& \text{Li}_4\left(\frac{1}{2}\right)-\frac{1}{8} \zeta (3) \log (2)+\frac{\pi ^4}{720}+\frac{\log ^4(2)}{24}\\ {\bf H}^{(1)}_4(1/2) &=& 2 \text{Li}_5\left(\frac{1}{2}\right)+\text{Li}_4\left(\frac{1}{2}\right) \log (2)+\frac{\zeta (5)}{32}-\frac{1}{36} \pi ^2 \left(3 \zeta (3)+\log ^3(2)\right)+\frac{1}{2} \zeta (3) \log ^2(2)+\frac{\log ^5(2)}{40}-\frac{1}{720} \pi ^4 \log (2) \end{eqnarray} Now, the question whether ${\bf H}^{(1)}_5(1/2)$ is expressible in terms of poly-logarithms is most likely answered negatively. Indeed I used the web interface http://wayback.cecm.sfu.ca/cgi-bin/EZFace/zetaform.cgi to find possible linear dependencies. The code

lindep([ zp(2,6)+zp(2,5,1), z(6), z(3)*z(3), z(3)*z(2)*log(2), z(3)*log(2)^3, z(2)*log(2)^4, z(2)^2*log(2)^2, z(5)*log(2), log(2)^6, zp(2,4)*log(2)^2, zp(2,5)*log(2), zp(2,6)])

being run for two different number of digits of precision gives completely different results which suggests that such linear dependency most likely does not exist.

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forcing the change of variable $x=\frac{1-y}{y}$, we have \begin{align} I=&\int_0^1\frac{\ln^4(1+x)\ln x}{x}\ dx=-\int_{1/2}^1\frac{\ln^5x}{x}\ dx-\underbrace{\int_{1/2}^1\frac{\ln^5x}{1-x}\ dx}_{IBP}+\int_{1/2}^1\frac{\ln^4x\ln(1-x)}{x(1-x)}\ dx\\ &=-\frac56\ln^62-5\int_{1/2}^1\frac{\ln(1-x)\ln^4x}{x}\ dx+\int_{1/2}^1\frac{\ln^4x\ln(1-x)}{x(1-x)}\ dx\\ &=-\frac56\ln^62+\sum_{n=1}^\infty\frac5n\int_{1/2}^1x^{n-1}\ln^4x\ dx-\sum_{n=1}^\infty H_n\int_{1/2}^1x^{n-1}\ln^4x\ dx\\ &=-\frac56\ln^62+\sum_{n=1}^\infty\left(\frac5n-H_n\right)\int_{1/2}^1x^{n-1}\ln^4x\ dx\\ &=-\frac56\ln^62+\sum_{n=1}^\infty\left(\frac5n-H_n\right)\left(\frac{24}{n^5}-\frac{24}{n^52^n}-\frac{24\ln2}{n^42^n}-\frac{12\ln^22}{n^32^n}-\frac{4\ln^32}{n^22^n}-\frac{\ln^42}{n2^n}\right)\\ &=-\frac56\ln^62+120\zeta(6)-120\operatorname{Li}_6\left(\frac12\right)-120\ln2\operatorname{Li}_5\left(\frac12\right)-60\ln^22\operatorname{Li}_4\left(\frac12\right)\\ &\quad-20\ln^32\operatorname{Li}_3\left(\frac12\right)-5\ln^42\operatorname{Li}_2\left(\frac12\right)-24\sum_{n=1}^\infty\frac{H_n}{n^5}+24\sum_{n=1}^\infty\frac{H_n}{n^52^n}+24\ln2\sum_{n=1}^\infty\frac{H_n}{n^42^n}\\ &\quad+12\ln^22\sum_{n=1}^\infty\frac{H_n}{n^32^n}+4\ln^32\sum_{n=1}^\infty\frac{H_n}{n^22^n}+\ln^42\sum_{n=1}^\infty\frac{H_n}{n2^n} \end{align} I was able here to prove: \begin{align} \sum_{n=1}^\infty \frac{H_n}{2^nn^3}=\operatorname{Li}_4\left(\frac12\right)+\frac18\zeta(4)-\frac18\ln2\zeta(3)+\frac1{24}\ln^42 \end{align} and here: \begin{align} \displaystyle\sum_{n=1}^{\infty}\frac{H_n}{2^n n^4}&=2\operatorname{Li_5}\left( \frac12\right)+\ln2\operatorname{Li_4}\left( \frac12\right)-\frac16\ln^32\zeta(2) +\frac12\ln^22\zeta(3)\\ &\quad-\frac18\ln2\zeta(4)- \frac12\zeta(2)\zeta(3)+\frac1{32}\zeta(5)+\frac1{40}\ln^52 \end{align} plugging these two sums along with the following well known values: $$\sum_{n=1}^\infty\frac{H_n}{n2^n}=\frac12\zeta(2)$$ $$\sum_{n=1}^\infty\frac{H_n}{n^22^n}=\zeta(3)-\frac12\ln2\zeta(2)$$ $$\sum_{n=1}^\infty\frac{H_n}{n^5}=\frac74\zeta(6)-\frac12\zeta^2(3)$$ $$\operatorname{Li_2}\left( \frac12\right) =\frac12\zeta(2)-\frac12\ln^22$$ $$\operatorname{Li_3}\left( \frac12\right)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$$ we obtain: $$I=-120\operatorname{Li}_6\left(\frac12\right)-72\ln2\operatorname{Li}_5\left(\frac12\right)-24\ln^22\operatorname{Li}_4\left(\frac12\right)+78\zeta(6)+\frac34\ln2\zeta(5)-\frac32\ln^22\zeta(4)-3\ln^32\zeta(3)+2\ln^42\zeta(2)+12\zeta^2(3)-12\ln2\zeta(2)\zeta(3)-\frac{17}{30}\ln^62+24\sum_{n=1}^\infty\frac{H_n}{n^52^n}$$

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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} &\bbox[#ffd,10px]{\ds{% \int_{0}^{1}{\ln^{4}\pars{1 + x}\ln\pars{x} \over x}\,\dd x}} \,\,\,\stackrel{1 + x\ \mapsto\ x}{=}\,\,\, \int_{1}^{2}{\ln^{4}\pars{x}\ln\pars{x - 1} \over x - 1}\,\dd x \\[5mm] \stackrel{x\ \mapsto\ 1/x}{=}\,\,\,& \int_{1}^{1/2}{\ln^{4}\pars{1/x}\ln\pars{1/x - 1} \over 1/x - 1} \,\pars{-\,{\dd x \over x^{2}}} = \int_{1/2}^{1}{\ln^{4}\pars{x}\ln\pars{1 - x} - \ln^{5}\pars{x} \over x\pars{1 - x}}\,\dd x \\[1cm] = &\ \underbrace{\int_{1/2}^{1}{\ln^{4}\pars{x}\ln\pars{1 - x} \over x}\,\dd x} _{\ds{\mc{I}_{1}}}\ -\ \underbrace{\int_{1/2}^{1}{\ln^{5}\pars{x} \over x}\,\dd x} _{\ds{=\ -\,{\ln^{6}\pars{2} \over 6}}}\ +\ \underbrace{\int_{1/2}^{1}{\ln^{4}\pars{x}\ln\pars{1 - x} \over 1 - x}\,\dd x} _{\ds{\mc{I}_{2}}} \\[1mm] &\ -\int_{1/2}^{1}{\ln^{5}\pars{x} \over 1 - x}\,\dd x \label{1}\tag{1} \end{align} Note that \begin{align} \int_{1/2}^{1}{\ln^{5}\pars{x} \over 1 - x}\,\dd x & = -\ln^{6}\pars{2} + 5\int_{1/2}^{1}{\ln^{4}\pars{x}\ln\pars{1 - x} \over x}\,\dd x = -\ln^{6}\pars{2} + 5\,\mc{I}_{1} \end{align} such that \eqref{1} becomes \begin{equation} \bbx{\bbox[#ffd,10px]{\ds{% \int_{0}^{1}{\ln^{4}\pars{1 + x}\ln\pars{x} \over x}\,\dd x}} = {7\ln^{6}\pars{2} \over 6} - 4\,\mc{I}_{1} + \mc{I}_{2}}\label{2}\tag{2} \end{equation}


$\ds{\Huge\mc{I}_{1}:\ ?}$. \begin{align} \mc{I}_{1} & \equiv \int_{1/2}^{1}{\ln^{4}\pars{x}\ln\pars{1 - x} \over x}\,\dd x = -\int_{1/2}^{1}\mrm{Li}_{2}'\pars{x}\ln^{4}\pars{x}\,\dd x \\[5mm] & = \mrm{Li}_{2}\pars{1 \over 2}\ln^{4}\pars{2} + 4\int_{1/2}^{1}\mrm{Li}_{3}'\pars{x}\ln^{3}\pars{x}\,\dd x \\[5mm] & = \mrm{Li}_{2}\pars{1 \over 2}\ln^{4}\pars{2} + 4\,\mrm{Li}_{3}\pars{1 \over 2}\ln^{3}\pars{2} - 12\int_{1/2}^{1}\mrm{Li}_{4}'\pars{x}\ln^{2}\pars{x}\,\dd x \\[5mm] & = \mrm{Li}_{2}\pars{1 \over 2}\ln^{4}\pars{2} + 4\,\mrm{Li}_{3}\pars{1 \over 2}\ln^{3}\pars{2} + 12\,\mrm{Li}_{4}\pars{1 \over 2}\ln^{2}\pars{2} + 24\int_{1/2}^{1}\mrm{Li}_{5}'\pars{x}\ln\pars{x}\,\dd x \\[1cm] & = \mrm{Li}_{2}\pars{1 \over 2}\ln^{4}\pars{2} + 4\,\mrm{Li}_{3}\pars{1 \over 2}\ln^{3}\pars{2} + 12\,\mrm{Li}_{4}\pars{1 \over 2}\ln^{2}\pars{2} + 24\,\mrm{Li}_{5}\pars{1 \over 2}\ln\pars{2} \\[1mm] & - 24\int_{1/2}^{1}\mrm{Li}_{6}'\pars{x}\,\dd x \\[1cm] & = \mrm{Li}_{2}\pars{1 \over 2}\ln^{4}\pars{2} + 4\,\mrm{Li}_{3}\pars{1 \over 2}\ln^{3}\pars{2} + 12\,\mrm{Li}_{4}\pars{1 \over 2}\ln^{2}\pars{2} + 24\,\mrm{Li}_{5}\pars{1 \over 2}\ln\pars{2} \\[1mm] & -24\,\mrm{Li}_{6}\pars{1} + 24\,\mrm{Li}_{6}\pars{1 \over 2} \end{align}

Note that $\ds{\mrm{Li}_{2}\pars{1 \over 2} = {\pi^{2} \over 12} - {\ln^{2}\pars{2} \over 2}}$ and $\ds{\mrm{Li}_{3}\pars{1 \over 2} = {\ln^{3}\pars{2} \over 6} - {\pi^{2}\ln\pars{2} \over 12} + {7\zeta\pars{3} \over 8}}$. Moreover, $\ds{\vphantom{\Huge A}\mrm{Li}_{6}\pars{1} = \zeta\pars{6} = {\pi^{6} \over 945}}$.

Then, $$ \begin{array}{|rcl|}\hline \mbox{}&&\\ \ds{\quad\mc{I}_{1}} & \ds{\equiv} & \ds{\int_{1/2}^{1}{\ln^{4}\pars{x}\ln\pars{1 - x} \over x}\,\dd x} \\[5mm] & \ds{=} & \ds{-\,{8\pi^{6} \over 315} - {\pi^{2}\ln^{4}\pars{2} \over 4} + {\ln^{6}\pars{2} \over 6} + 12\ln^{2}\pars{2}\,\mrm{Li}_{4}\pars{1 \over 2} + 24\ln\pars{2}\,\mrm{Li}_{5}\pars{1 \over 2}\quad} \\[1mm] && \ds{% + 24\,\mrm{Li}_{6}\pars{1 \over 2} + {7\ln^{3}\pars{2}\,\zeta\pars{3} \over 2} \approx -0.0269} \\ \mbox{}&& \\ \hline \end{array} $$


$\ds{\Huge\mc{I}_{2}:\ ?}$. This one isn't trivial at all. An attempt is given by \begin{align} \mc{I}_{2} & \equiv \int_{1/2}^{1}{\ln^{4}\pars{x}\ln\pars{1 - x} \over 1 - x}\,\dd x \,\,\,\stackrel{x\ \mapsto\ 1 - x}{=}\,\,\, \int_{0}^{1/2}{\ln^{4}\pars{1 - x}\ln\pars{x} \over x}\,\dd x \\[5mm] & = \left.\partiald[4]{}{\mu}\partiald{}{\nu} \int_{0}^{1/2}\bracks{\pars{1 - x}^{\mu} - 1}x^{\nu - 1} \,\dd x\,\right\vert_{\ \mu\ =\ 0\,,\ \nu\ =\ 0^{+}} \\[5mm] & = \partiald[4]{}{\mu}\partiald{}{\nu} \bracks{\int_{0}^{1/2}\pars{1 - x}^{\mu}\,x^{\nu - 1} \,\dd x - {1 \over 2^{\nu}\nu}}_{\ \mu\ =\ 0\,,\ \nu\ =\ 0^{+}} \\[5mm] & = \partiald[4]{}{\mu}\partiald{}{\nu} \bracks{\mrm{B}\pars{{1 \over 2},\nu,1 + \mu} - {1 \over 2^{\nu}\nu}}_{\ \mu\ =\ 0\,,\ \nu\ =\ 0^{+}} \end{align}

$\ds{\mrm{B}}$ is the Incomplete Beta Function. It'll continue$\ldots$

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  • $\begingroup$ $I_2$ will be in terms of $\displaystyle \sum_{n=1}^\infty\frac{H_n}{n^52^n}$ which has no closed form. $\endgroup$ – Ali Shather Jun 2 at 21:03

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