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I need your help in evaluating the following integral in closed form. $\displaystyle\int\limits_{0.5}^{1} \frac{\mathrm{Li}_{2}\left(x\right)\ln\left(2x - 1\right)}{x}\,\mathrm{d}x$

Since the function is singular at $x = 0.5$, we are looking for Principal Value. The integral is finite and was evaluated numerically. I expect the closed form result to contain $\,\mathrm{Li}_{3}$ and $\,\mathrm{Li}_{2}$.

Thanks

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    $\begingroup$ What have you tried? Any Taylor expansions? Note that $\ln(2x-1)=\ln(1-1/2x)+\ln(x)+\ln(2)$ $\endgroup$ – Simply Beautiful Art Aug 29 '17 at 13:43
  • $\begingroup$ Could you explain why, in your opinion, a closed form exists and would contain $Li_3$ and $Li_2$ ? This would be of interest to me. $\endgroup$ – Claude Leibovici Aug 29 '17 at 14:04
  • $\begingroup$ Hi Claude, As a starting point, why don't you ignore my ``speculation" regarding the final form, and attempt to solve it like any new unsolved integral. $\endgroup$ – Hmath Aug 29 '17 at 16:52
  • $\begingroup$ This is simply $-I\bigg(\dfrac12\bigg),$ where $I(a)=\displaystyle\int_a^1\frac{\text{Li}_3(x)-\text{Li}_3(a)}{x-a}~dx.$ $\endgroup$ – Lucian Sep 8 '17 at 21:23
  • $\begingroup$ It resembles Frullani's integral. $\endgroup$ – Lucian Sep 8 '17 at 21:31
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I will be using next integral $$ \operatorname{Lv}_{n}(x,\alpha) =\int\limits_{0}^{x} \dfrac{\ln^n t}{1-\alpha t}\,\mathrm{d}t = \dfrac{n!}{\alpha}\sum\limits_{k\,=\,0}^{n}\dfrac{(-1)^k}{\left(n-k\right)!}\ln^{n-k}x\operatorname{Li}_{k+1}(\alpha x) $$ with $x \geq 0$, $n \in \mathbb{N}$ and $\alpha$ such that $$ \alpha x < 1 \text{ or } \alpha x = 1\ \wedge\ x = 1$$ If anywhere $\operatorname{Lv}$ will appear I'll omit calculations of $\operatorname{Li}$ constants. Let \begin{align*}\mathfrak{I} &= \int\limits_{1/2}^{1}\dfrac{\ln\left(2x-1\right)}{x}\operatorname{Li}_2(x)\,\mathrm{d}x \\ &= \int\limits_{0}^{1}\dfrac{\ln x}{1+x}\operatorname{Li}_2\left(\dfrac{1+x}{2}\right)\,\mathrm{d}x \\ &= \operatorname{Li}_2\left(\dfrac{1+x}{2}\right)\Big(\ln x\ln\left(1+x\right) + \operatorname{Li}_2(-x)\Big)\Bigg\vert_{0}^{1}+\int\limits_{0}^{1}\dfrac{\ln x\ln\left(1+x\right) + \operatorname{Li}_2(-x)}{1+x}\ln \left(\dfrac{1-x}{2}\right)\,\mathrm{d}x \\ &= \operatorname{Li}_2(1)\operatorname{Li}_2(-1) + \underbrace{\int\limits_{0}^{1}\dfrac{\ln\left(1-x\right)\ln x\ln\left(1+x\right)}{1+x}\,\mathrm{d}x}_{\mathfrak{I}_1}+\underbrace{\int\limits_{0}^{1}\dfrac{\operatorname{Li}_2(-x)\ln\left(1-x\right)}{1+x}\,\mathrm{d}x}_{\mathfrak{I}_2}-\phantom{a}\\ &-\ln 2\underbrace{\int\limits_{0}^{1}\dfrac{1}{1+x}\left(\int\limits_{0}^{x}\dfrac{\ln t}{1+t}\,\mathrm{d}t\right)\,\mathrm{d}x}_{\mathfrak{I}_3} \end{align*}

$\mathfrak{I}_1:$

Make substitution $x\rightarrow \dfrac{1-x}{1+x}$ \begin{align*} \mathfrak{I}_1 &= \int\limits_{0}^{1} \dfrac{1}{1+x}\ln\left(\dfrac{2}{1+x}\right)\ln\left(\dfrac{1-x}{1+x}\right)\ln\left(\dfrac{2x}{1+x}\right)\,\mathrm{d}x \\ &= \ln 2 \int\limits_{0}^{1} \dfrac{\ln\left(1-x\right)\ln x}{1+x}\,\mathrm{d}x - \ln 2 \int\limits_{0}^{1} \dfrac{\ln\left(1+x\right)\ln x}{1+x}\,\mathrm{d}x - \mathfrak{I}_1 + \int\limits_{0}^{1} \dfrac{\ln^2\left(1+x\right)\ln x}{1+x}\,\mathrm{d}x + \phantom{a} \\ &\,+ \color{red}{\int\limits_{0}^{1} \dfrac{1}{1+x}\ln^2\left(\dfrac{2}{1+x}\right)\ln\left(\dfrac{1-x}{1+x}\right)\,\mathrm{d}x} \\ &= \ln 2 \underbrace{\int\limits_{0}^{1} \dfrac{\ln\left(1-x\right)\ln x}{1+x}\,\mathrm{d}x}_{\mathfrak{I}_{1,1}} - \ln 2 \int\limits_{0}^{1} \dfrac{\ln\left(1+x\right)\ln x}{1+x}\,\mathrm{d}x - \mathfrak{I}_1 + \color{red}{2}\int\limits_{0}^{1} \dfrac{\ln^2\left(1+x\right)\ln x}{1+x}\,\mathrm{d}x \\ &= \dfrac{1}{2}\mathfrak{I}_{1,1}\ln 2 - \dfrac{1}{4}\ln 2 \ln^2\left(1+x\right)\ln x\Bigg\vert_{0}^{1} +\dfrac{1}{4}\ln 2 \int\limits_{0}^{1} \dfrac{\ln^2\left(1+x\right)}{x}\,\mathrm{d}x+\dfrac{1}{3}\ln^3\left(1+x\right)\ln x\Bigg\vert_{0}^{1} - \phantom{a} \\ &\ - \dfrac{1}{3}\int\limits_{0}^{1} \dfrac{\ln^3\left(1+x\right)}{x}\,\mathrm{d}x \\ &= \dfrac{1}{2}\mathfrak{I}_{1,1}\ln 2 + \dfrac{1}{4}\ln 2 \int\limits_{1}^{2} \dfrac{\ln^2 x}{x-1}\,\mathrm{d}x - \dfrac{1}{3}\int\limits_{1}^{2} \dfrac{\ln^3 x}{x-1}\,\mathrm{d}x \\ &= \dfrac{1}{2}\mathfrak{I}_{1,1}\ln 2 + \dfrac{1}{4}\ln 2 \int\limits_{1/2}^{1} \dfrac{\ln^2 x}{x\left(1-x\right)}\,\mathrm{d}x + \dfrac{1}{3}\int\limits_{1/2}^{1} \dfrac{\ln^3 x}{x\left(1-x\right)}\,\mathrm{d}x \\ &= \dfrac{1}{2}\mathfrak{I}_{1,1}\ln 2 + \dfrac{1}{4}\ln 2\left(\dfrac{1}{3}\ln^3 2 + \operatorname{Lv}_{2}(1,1)-\operatorname{Lv}_{2}\left(\dfrac{1}{2},1\right)\right) + \dfrac{1}{3}\left(-\dfrac{1}{4}\ln^4 2 + \operatorname{Lv}_{3}(1,1)-\operatorname{Lv}_{3}\left(\dfrac{1}{2},1\right)\right) \\ &= \dfrac{1}{2}\mathfrak{I}_{1,1}\ln 2 + 2\operatorname{Li}_4\left(\dfrac{1}{2}\right) + \dfrac{29}{16}\zeta(3)\ln 2 - \dfrac{1}{45}\pi^4 + \dfrac{1}{12}\ln^4 2 - \dfrac{1}{12}\pi^2\ln^2 2 \end{align*}

$\mathfrak{I}_{1,1}:$

Use identity $$ xy = \dfrac{1}{2}x^2 + \dfrac{1}{2}y^2 - \dfrac{1}{2}\left(x-y\right)^2$$ to have \begin{align*} \mathfrak{I}_{1,1} &= \dfrac{1}{2}\int\limits_{0}^{1} \dfrac{\ln^2 x}{1+x}\,\mathrm{d}x + \dfrac{1}{2}\int\limits_{0}^{1} \dfrac{\ln^2 \left(1-x\right)}{1+x}\,\mathrm{d}x - \dfrac{1}{2}\int\limits_{0}^{1} \dfrac{1}{1+x}\ln^2\left(\dfrac{x}{1-x}\right)\,\mathrm{d}x \\ &= \dfrac{1}{2}\operatorname{Lv}_2(1,-1)+\dfrac{1}{2}\int\limits_{0}^{1} \dfrac{\ln^2 x}{2-x}\,\mathrm{d}x-\dfrac{1}{2}\int\limits_{0}^{\infty} \dfrac{\ln^2 x}{\left(1+2x\right)\left(1+x\right)}\,\mathrm{d}x \\ &= \dfrac{1}{2}\operatorname{Lv}_2(1,-1)+\dfrac{1}{4}\operatorname{Lv}_2\left(1,\dfrac{1}{2}\right)-\dfrac{1}{2}\int\limits_{0}^{1} \dfrac{\ln^2 x}{\left(1+2x\right)\left(1+x\right)}\,\mathrm{d}x-\dfrac{1}{2}\int\limits_{0}^{1} \dfrac{\ln^2 x}{\left(2+x\right)\left(1+x\right)}\,\mathrm{d}x \\ &= \dfrac{1}{2}\operatorname{Lv}_2(1,-1)+\dfrac{1}{4}\operatorname{Lv}_2\left(1,\dfrac{1}{2}\right)-\int\limits_{0}^{1} \dfrac{\ln^2 x}{1+2x}\,\mathrm{d}x+\dfrac{1}{2}\int\limits_{0}^{1} \dfrac{\ln^2 x}{2+x}\,\mathrm{d}x \\ &= \dfrac{1}{2}\operatorname{Lv}_2(1,-1)+\dfrac{1}{4}\operatorname{Lv}_2\left(1,\dfrac{1}{2}\right)-\operatorname{Lv}_2(1,-2)+\dfrac{1}{4}\operatorname{Lv}_2\left(1,-\dfrac{1}{2}\right) \\ &= \dfrac{13}{8}\zeta(3)-\dfrac{1}{4}\pi^2\ln 2 \end{align*} So

$$\mathfrak{I}_1 = 2\operatorname{Li}_4\left(\dfrac{1}{2}\right)+\dfrac{21}{8}\zeta(3)\ln 2 - \dfrac{1}{45}\pi^4 + \dfrac{1}{12}\ln^4 2 - \dfrac{5}{24}\pi^2\ln^2 2$$


$\mathfrak{I}_{2}:$

Use the same identity as for $\mathfrak{I}_{1,1}$

\begin{align*} -\mathfrak{I}_2 &= \int\limits_{0}^{1} \dfrac{\ln\left(1-x\right)}{1+x}\left(\int\limits_{0}^{1}\dfrac{\ln\left(1+xt\right)}{t}\,\mathrm{d}t\right)\,\mathrm{d}x \\ &= \dfrac{1}{2}\int\limits_{0}^{1} \dfrac{1}{1+x}\left(\int\limits_{0}^{1}\dfrac{\ln^2\left(1+xt\right)}{t}\,\mathrm{d}t\right)\,\mathrm{d}x + \dfrac{1}{2}\int\limits_{0}^{1}\int\limits_{0}^{1} \dfrac{1}{\left(1+x\right)t}\left(\ln^2\left(1-x\right)-\ln^2\left(\dfrac{1-x}{1+xt}\right)\right)\,\mathrm{d}t\,\mathrm{d}x \\ &= \dfrac{1}{2}\int\limits_{0}^{1} \dfrac{1}{1+x}\left(\int\limits_{0}^{x}\dfrac{\ln^2\left(1+t\right)}{t}\,\mathrm{d}t\right)\,\mathrm{d}x + \phantom{a} \\ &+ \dfrac{1}{2}\int\limits_{0}^{1} \dfrac{1}{t}\left(\int\limits_{0}^{1}\dfrac{\ln^2\left(1-x\right)}{1+x}\,\mathrm{d}x-\int\limits_{0}^{1}\dfrac{1}{1+x}\ln^2\left(\dfrac{1-x}{1+xt}\right)\,\mathrm{d}x\right)\,\mathrm{d}t \\ &= \dfrac{1}{2}\left.\ln\left(1+x\right)\int\limits_{0}^{x}\dfrac{\ln^2\left(1+t\right)}{t}\,\mathrm{d}t\right\vert_{0}^{1}-\dfrac{1}{2}\int\limits_{0}^{1}\dfrac{\ln^3\left(1+x\right)}{x}\,\mathrm{d}x + \phantom{a} \\ &+ \dfrac{1}{2}\int\limits_{0}^{1} \dfrac{1}{t}\left(\int\limits_{0}^{1}\dfrac{\ln^2 x}{2-x}\,\mathrm{d}x-\left(1+t\right)\int\limits_{0}^{1}\dfrac{\ln^2 x}{\left(1+xt\right)\left(2-x\left(1-t\right)\right)}\,\mathrm{d}x\right)\,\mathrm{d}t \\ &= \dfrac{1}{2}\ln 2\left(\dfrac{1}{3}\ln^3 2 + \operatorname{Lv}_2(1,1)-\operatorname{Lv}_2\left(\dfrac{1}{2},1\right)\right)+\dfrac{1}{2}\left(-\dfrac{1}{4}\ln^4 2+\operatorname{Lv}_3(1,1)-\operatorname{Lv}_3\left(\dfrac{1}{2},1\right)\right)+\phantom{a} \\ &+ \dfrac{1}{2}\int\limits_{0}^{1} \dfrac{1}{t}\left(\dfrac{1}{2}\operatorname{Lv}_2\left(1,\dfrac{1}{2}\right)-\int\limits_{0}^{1} \ln^2 x\left(\dfrac{t}{1+xt}+\dfrac{1-t}{2-x\left(1-t\right)}\right)\,\mathrm{d}x\right)\,\mathrm{d}t \\ &= 3\operatorname{Li}_4\left(\dfrac{1}{2}\right)+\dfrac{11}{4}\zeta(3)\ln 2 - \dfrac{1}{30}\pi^4 + \dfrac{1}{8}\ln^4 2 - \dfrac{1}{8}\pi^2\ln^2 2+\phantom{a} \\ &+ \dfrac{1}{2}\underbrace{\int\limits_{0}^{1} \dfrac{1}{t}\left(\dfrac{1}{2}\operatorname{Lv}_2\left(1,\dfrac{1}{2}\right)-t\operatorname{Lv}_2(1,-t)-\dfrac{1-t}{2}\operatorname{Lv}_2\left(1, \dfrac{1-t}{2}\right)\right)\,\mathrm{d}t}_{\mathfrak{I}_{2,1}} \end{align*} Note that $$ \operatorname{Lv}_2(1,\alpha) = \dfrac{2}{\alpha}\operatorname{Li}_3(\alpha)$$ So $\mathfrak{I}_{2,1}$ can be simplified to \begin{align*} \mathfrak{I}_{2,1} &= \int\limits_{0}^{1} \dfrac{1}{t}\left(2\operatorname{Li}_3\left(\dfrac{1}{2}\right)+2\operatorname{Li}_3(-t)-2\operatorname{Li}_3\left(\dfrac{1-t}{2}\right)\right)\,\mathrm{d}t \\ &= 2\int\limits_{0}^{1} \dfrac{\operatorname{Li}_3(-t)}{t}\,\mathrm{d}t+2\int\limits_{0}^{1} \dfrac{1}{t}\left(\operatorname{Li}_3\left(\dfrac{1}{2}\right)-\operatorname{Li}_3\left(\dfrac{1-t}{2}\right)\right)\,\mathrm{d}t \\ &= 2\operatorname{Li}_4(-1)+2\sum\limits_{n=1}^{\infty}\dfrac{1}{n^32^n}\int\limits_{0}^{1} \dfrac{1-t^n}{1-t}\,\mathrm{d}t \\ &= 2\operatorname{Li}_4(-1)+2\sum\limits_{n=1}^{\infty}\dfrac{\mathcal{H}_n}{n^32^n} \\ &= 2\operatorname{Li}_4\left(\dfrac{1}{2}\right) - \dfrac{1}{4}\zeta(3)\ln 2 - \dfrac{1}{60}\pi^4 + \dfrac{1}{12}\ln^4 2 \end{align*} Closed form for series above you can find here. So

$$\mathfrak{I}_2 = -4\operatorname{Li}_4\left(\dfrac{1}{2}\right)-\dfrac{21}{8}\zeta(3)\ln 2 + \dfrac{1}{24}\pi^4-\dfrac{1}{6}\ln^4 2 + \dfrac{1}{8}\pi^2\ln^2 2$$


$\mathfrak{I}_3:$

\begin{align*} \mathfrak{I}_3 &= \ln\left(1+x\right)\Big(\ln x\ln\left(1+x\right) + \operatorname{Li}_2(-x)\Big)\Bigg\vert_{0}^{1} - \int\limits_{0}^{1}\dfrac{\ln\left(1+x\right)\ln x}{1+x}\,\mathrm{d}x \\ &= \ln 2\operatorname{Li}_2(-1) - \dfrac{1}{2}\ln^2\left(1+x\right)\ln x\Bigg\vert_{0}^{1}+\dfrac{1}{2}\int\limits_{0}^{1} \dfrac{\ln^2\left(1+x\right)}{x}\,\mathrm{d}x \\ &= \ln 2\operatorname{Li}_2(-1) + \dfrac{1}{2}\left(\dfrac{1}{3}\ln^3 2 + \operatorname{Lv}_2(1,1) - \operatorname{Lv}_2\left(\dfrac{1}{2},1\right)\right) \\ &= \dfrac{1}{8}\zeta(3)-\dfrac{1}{12}\pi^2\ln 2 \end{align*} So

$$\mathfrak{I}_3 = \dfrac{1}{8}\zeta(3)-\dfrac{1}{12}\pi^2\ln 2$$


Collecting all parts together we have that \begin{align*} \mathfrak{I} &= \operatorname{Li}_2(1)\operatorname{Li}_2(-1)+\mathfrak{I}_1+\mathfrak{I}_2-\mathfrak{I}_3\ln 2 \\ &= -2\operatorname{Li}_4\left(\dfrac{1}{2}\right)-\dfrac{1}{8}\zeta(3)\ln 2 + \dfrac{1}{180}\pi^4 - \dfrac{1}{12}\ln^4 2 \end{align*}

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  • $\begingroup$ very nice solution (+1). $\endgroup$ – Ali Shadhar Mar 26 '20 at 21:19
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This provisional answer is an horribly inelegant conjecture with no proof attached

$$\int\limits_{0.5}^{1}\frac{Li_2(x) \ln(2x-1)}{x} dx =-\frac{1}{2}\sum_{n=1}^{\infty}\frac{\sum_{k=0}^{n-1} \frac{\binom{n-1}{k}}{(k+1)^2}}{ \sum_{k=0}^n (k (2 k-1)) \binom{n}{k}}\tag1$$

I have checked this solution using $m=100$ approximation

$$\int_{\frac{1}{2}}^1 \frac{ \left(\sum _{k=1}^{m} \frac{x^k}{k^2}\right)\log (2 x-1)}{x} \, dx =-\frac{1}{2}\sum_{n=1}^{m}\frac{\sum_{k=0}^{n-1} \frac{\binom{n-1}{k}}{(k+1)^2}}{ \sum_{k=0}^n (k (2 k-1)) \binom{n}{k}}$$

The numerator summation arises from the number sequence given here (binomial transform of $1/(k+1)^2$: that is 1, 5/4, 29/18, 103/48, 887/300, 1517/360, etc.) and the denominator summation arises from the number sequence given here (the binomial transform of the hexagonal numbers)

Hopefully someone can make a little more sense of this than I have.

Later Edit: In working to understand user90369's answer to this question I found these identities using Mathematica

$$Li_n(\frac{2}{m})-Li_n(\frac{1}{m})=\sum _{k=1}^{\infty } \frac{1}{m^k k^{n-1}}\sum _{v=1}^k \binom{k-1}{v-1}\frac{1}{v}$$

in the case of $m=2$ $$\zeta(n)-Li_n(\frac{1}{2})=\sum _{k=1}^{\infty } \frac{1}{2^k k^{n-1}}\sum _{v=1}^k \binom{k-1}{v-1}\frac{1}{v}$$

Added Later Still: More Trivial Identities

$$Li_2(\frac{1}{2})=\sum\limits_{k=1}^\infty \frac{1}{k^3 2^k}\sum\limits_{v=1}^1 {\binom k v} \frac{1}{v}$$

$$Li_4(\frac{1}{2})=\sum\limits_{k=1}^\infty \frac{1}{k^3 2^k}\sum\limits_{v=k}^k {\binom k v} \frac{1}{v}$$

$$\int\limits_{0.5}^{1}\frac{Li_2(x) \ln(2x-1)}{x} dx=-Li_2(\frac{1}{2})-Li_4(\frac{1}{2})-\sum\limits_{k=1}^\infty \frac{1}{k^3 2^k}\sum\limits_{v=2}^{k-1} {\binom k v} \frac{1}{v}$$

Since $$\sum\limits_{k=1}^{\infty} \frac{1}{k^3 2^k}\sum\limits_{v=1}^{k} {\binom k v} \frac{1}{v}=\sum _{v=1}^{\infty} \frac{1}{v}\sum _{k=1}^{\infty} \frac{1}{k^{3} 2^k}\binom{k}{v}$$

There is a brute force pattern matching approach that I have found, using

$$S_a=\sum\limits_{k=1}^{\infty} \frac{1}{k^3 2^k}\sum\limits_{v=a}^{a} {\binom k v} \frac{1}{v}=\sum _{v=a}^a \frac{1}{v}\sum _{k=1}^{\infty} \frac{1}{k^{3} 2^k}\binom{k}{v}$$

where $$\int\limits_{0.5}^{1}\frac{Li_2(x) \ln(2x-1)}{x} dx= -\sum_{a=1}^{\infty} S_a$$

From Mathematica $$S_1=\frac{\pi ^2}{12}-\frac{\log ^2(2)}{2}=Li_2(\frac{1}{2})$$ $$S_2=1/48 \left(-\pi^2 + 12 \log(2) + 6 \log^2(2) \right)$$ $$S_3=\frac{1}{108} \left(6+\pi ^2-18 \log (2)-6 \log ^2(2)\right) $$ $$S_4=\frac{1}{192} \left(-8-\pi ^2+22 \log (2)+6 \log ^2(2)\right)$$

and so on. From $S_2$ onwards the general term as far as I have been able to determine is

$$S_a=(-1)^{k-1}C_a+(-1)^{k-1}\frac{\pi^2}{12a^2}+(-1)^{k}\frac{H_{a-1}\log(2)}{a^2}+(-1)^{k}\frac{\log^2(2)}{2a^2}$$

where $H_a$ is the Harmonic Number. I haven't been able to determine the pattern for the rational term, $C_a$ [$0$,$\frac{6}{(12\times3^2)}$,$\frac{8}{(12\times4^2)}$,$\frac{21}{2(12\times5^2)}$,$\frac{119}{10(12\times6^2)}$,$\frac{202}{15(12\times7^2)}$,$\frac{1525}{105(12\times8^2)}$,...]. Since the last three sum up to known closed forms it would be unfortunate if the rational term summation didn't do the same.

Once again I hope someone can make a little more sense of this than I have.

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  • $\begingroup$ Nice idea. (+1) :-D But it seems to be that the OP is not any more interested in his own question. :-( $\endgroup$ – user90369 Sep 4 '17 at 14:20
  • $\begingroup$ I agree the OP seems to have lost interest. I might raise a new question in regards to determining a pattern for the rational term. $\endgroup$ – James Arathoon Sep 4 '17 at 14:57
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$\displaystyle \int\limits_{0.5}^1 \frac{Li_2(x)\ln(2x-1)}{x}dx=$

$\displaystyle =\sum\limits_{k=1}^\infty \frac{1}{k^2 2^k}\sum\limits_{v=0}^{k-1} {\binom {k-1} v} \lim\limits_{h\to 0}\frac{1}{h}\left(\frac{(2x-1)^{v+h+1}}{v+h+1}-\frac{(2x-1)^{v+1}}{v+1}\right)|_{0.5}^1$

$\displaystyle =-\sum\limits_{k=1}^\infty \frac{1}{k^3 2^k}\sum\limits_{v=1}^k {\binom k v} \frac{1}{v} = -\int\limits_0^1 \frac{Li_3(\frac{x+1}{2})-Li_3(\frac{1}{2})}{x}dx $

First note:

Be $\,\displaystyle H_k(x):=x\int\limits_0^1 \frac{(xt)^k-1}{xt-1}dt=\sum\limits_{v=1}^k \frac{x^v}{v}$ . $\,$ It's $\enspace\displaystyle \sum\limits_{v=1}^k {\binom k v} \frac{1}{v}=H_k(2)-H_k(1)$ .

Second note:

We can define e.g. $\,\displaystyle Fi_n(x):=\int\limits_0^{1-x}\frac{Li_n(t+x)-Li_n(x)}{t}dt\,$ for $\,|x|\leq 1\,$ .

Then it's $\,\displaystyle \int\limits_{0.5}^1 \frac{Li_2(x)\ln(2x-1)}{x}dx=-Fi_3(\frac{1}{2})\,$ .

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  • $\begingroup$ I did NOT loose interest in my question! The funny thing is I am evaluating this integral because I want to evaluate an infinite sum involving binomial coefficients! You have just brought me back to binomial coefficients... Definitely not the approach I was looking for $\endgroup$ – Hmath Sep 4 '17 at 17:33
  • $\begingroup$ Good to hear. I must add that you chose to hide the context behind your question. I gave a +1 to @SimplyBeautifulArt's comment "What have you tried?" If you had added context earlier (including what approach you were looking for) you could perhaps have attracted more interest and up-voting to what seems like a very interesting problem. $\endgroup$ – James Arathoon Sep 4 '17 at 18:15
  • $\begingroup$ @Hmath : I agree with James Arathoon, it's indeed not possible to understand what you are looking for (beside a "closed form", which is obviously not possible). And: $H_k(x)$ is without any binomial coefficients. $\endgroup$ – user90369 Sep 4 '17 at 19:39
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Start with subbing $ 2x-1\to x$ then integrate by parts we have

$$I=-\frac54\zeta(4)+\int_0^1\frac{\text{Li}_2(-x)}{1+x}\ln\left(\frac{1-x}{2}\right)\ dx+\int_0^1\frac{\ln(x)\ln(1+x)}{1+x}\ln\left(\frac{1-x}{2}\right)\ dx$$

$$=-\frac54\zeta(4)+\int_0^1\frac{\text{Li}_2(-x)\ln(1-x)}{1+x}\ dx+\int_0^1\frac{\ln(x)\ln(1+x)\ln(1-x)}{1+x}\ dx$$

$$-\ln(2)\int_0^1\frac{\ln(x)}{1+x}[\text{Li}_2(-x)+\ln(x)\ln(1+x)]\ dx$$

$$=-\frac54\zeta(4)+A+B-C$$

For $A$, use

$$\frac{\text{Li}_2(-x)}{1+x}=\sum_{n=1}^\infty (-1)^{n-1}H_{n-1}^{(2)}x^{n-1}\tag1$$

$$A=\sum_{n=1}^\infty (-1)^{n-1}H_{n-1}^{(2)}\int_0^1x^{n-1}\ln(1-x) \ dx=\sum_{n=1}^\infty (-1)^{n-1}H_{n-1}^{(2)}\left(-\frac{H_n}{n}\right)$$

$$=\sum_{n=1}^\infty \frac{(-1)^nH_{n}^{(2)}H_n}{n}-\sum_{n=1}^\infty \frac{(-1)^nH_n}{n^3}$$

For B, use

$$\frac{\ln(1+x)}{1+x}=\sum_{n=1}^\infty (-1)^{n}H_{n-1} x^{n-1}\tag2$$

$$B=\sum_{n=1}^\infty (-1)^{n}H_{n-1} \int_0^1x^{n-1}\ln(x)\ln(1-x)\ dx=\sum_{n=1}^\infty (-1)^{n}H_{n-1}\left(\frac{H_n}{n^2}+\frac{H_n^{(2)}}{n}-\frac{\zeta(2)}{n}\right)$$

$$=\sum_{n=1}^\infty \frac{(-1)^{n}H_{n-1}H_n}{n^2}+\sum_{n=1}^\infty \frac{(-1)^{n}H_{n-1}H_n^{(2)}}{n}-\zeta(2)\sum_{n=1}^\infty \frac{(-1)^{n}H_{n-1}}{n}$$

$$=\sum_{n=1}^\infty \frac{(-1)^{n}H_n^2}{n^2}-\sum_{n=1}^\infty \frac{(-1)^{n}H_n}{n^3}+\sum_{n=1}^\infty \frac{(-1)^{n}H_{n}H_n^{(2)}}{n}-\sum_{n=1}^\infty \frac{(-1)^{n}H_n^{(2)}}{n^2}-\zeta(2)\sum_{n=1}^\infty \frac{(-1)^{n}H_{n-1}}{n}$$

Combine the results of A and B we get

$$\small{A+B=\sum_{n=1}^\infty \frac{(-1)^{n}H_n^2}{n^2}-2\sum_{n=1}^\infty \frac{(-1)^{n}H_n}{n^3}+2\sum_{n=1}^\infty \frac{(-1)^{n}H_{n}H_n^{(2)}}{n}-\sum_{n=1}^\infty \frac{(-1)^{n}H_n^{(2)}}{n^2}-\zeta(2)\sum_{n=1}^\infty \frac{(-1)^{n}H_{n-1}}{n}}$$

The last sum can be easily obtained by integrating both sides of $(2)$, $\Longrightarrow \sum_{n=1}^\infty \frac{(-1)^{n}H_{n-1}}{n}=\frac12\ln^2(2)$

For the rest of the sums, they are already calculated:

$$\sum_{n=1}^\infty \frac{(-1)^{n}H_n}{n^3}=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42\tag3$$

$$\sum_{n=1}^{\infty}\frac{(-1)^nH_n^{(2)}}{n^2}=-4\operatorname{Li}_4\left(\frac12\right)+\frac{51}{16}\zeta(4)-\frac72\ln2\zeta(3)+\ln^22\zeta(2)-\frac16\ln^42\tag4$$

$$\sum_{n=1}^{\infty}\frac{(-1)^nH_n^2}{n^2}=2\operatorname{Li}_4\left(\frac12\right)-\frac{41}{16}\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac1{12}\ln^42\tag5$$

where $(3)$ can be found here and $(4)$ and $(5)$ can be found here.

Integral $C$ was nicely calculated by @Aknas ( check integral $\Im_3$ in his solution above) where he used

$$\text{Li}_2(x)+\ln(x)\ln(1+x)=\int_0^x\frac{\ln(t)}{1+t}\ dt$$

$$C= \dfrac{1}{8}\zeta(3)-\dfrac{1}{2}\ln(2)\zeta(2)$$

Combine all resuts of $A$, $B$ and $C$ we obtain that

$$I=-2\operatorname{Li}_4\left(\dfrac{1}{2}\right)-\dfrac{1}{8}\ln(2)\zeta(3) + \dfrac{1}{2}\zeta(4)- \dfrac{1}{12}\ln^4(2)$$

$\endgroup$

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