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How to prove without using Euler sums that

$$I=\int_0^1\frac{\ln x\ln(1+x)}{1-x}\ dx=\zeta(3)-\frac32\ln2\zeta(2)$$

where $\zeta$ is the Riemann zeta function.

We can relate this integral to some Euler sum as follows:

\begin{align} I&=-\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^1\frac{x^n\ln x}{1-x}\ dx\\ &=-\sum_{n=1}^\infty\frac{(-1)^n}{n}(H_n^{(2)}-\zeta(2))\\ &=-\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n}-\ln2\zeta(2) \end{align}

Also the integral $I$ can be related to $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^2}$. So I am looking for a different way to evaluate $I$ besides using these two sums.

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\begin{align}J&=\int_0^1 \frac{\ln x\ln(1+x)}{1-x}\,dx\end{align} Always the same story...

For $x\in [0;1]$ define the function $R$ by,\begin{align}R(x)&=\int_0^x \frac{\ln t}{1-t}\,dt\\ &=\int_0^1 \frac{x\ln(tx)}{1-tx}\,dt\\ J&=\Big[R(x)\ln(1+x)\Big]_0^1-\int_0^1 \frac{R(x)}{(1+x)} dx\\ &=-\zeta(2)\ln 2-\int_0^1 \int_0^1 \frac{x\ln(tx)}{(1-tx)(1+x)}\,dt\,dx\\ &=-\zeta(2)\ln 2-\int_0^1 \left(\int_0^1 \frac{x\ln t}{(1-tx)(1+x)}\,dx\right)\,dt-\int_0^1 \left(\int_0^1 \frac{x\ln x}{(1-tx)(1+x)}\,dt\right)\,dx\\ &=-\zeta(2)\ln 2+\int_0^1\left[\frac{\ln(1-tx)}{t(1+t)}+\frac{\ln(1+x)}{1+t}\right]_{x=0}^{x=1}\ln t\,dt+\int_0^1\left[\frac{\ln(1-tx)}{1+x}\right]_{t=0}^{t=1}\ln x\,dx\\ &=-\zeta(2)\ln 2+\int_0^1 \frac{\ln(1-t)\ln t}{t(1+t)}\,dt+\ln 2\int_0^1 \frac{\ln t}{1+t}\,dt+\int_0^1 \frac{\ln(1-x)\ln x}{1+x}\,dx\\ &=-\zeta(2)\ln 2+\int_0^1 \frac{\ln(1-t)\ln t}{t}\,dt+\ln 2\int_0^1 \frac{\ln t}{1+t}\,dt\\ &=-\zeta(2)\ln 2+\frac{1}{2}\left(\Big[\ln^2 x\ln(1-x)\Big]+\int_0^1 \frac{\ln^2 t }{1-t}\,dt\right)+\ln 2\left(\int_0^1 \frac{\ln t}{1-t}\,dt-\int_0^1 \frac{2t\ln t}{1-t^2}\,dt\right) \end{align} In the last integral perform the change of variable $y=t^2$, \begin{align}J&=-\zeta(2)\ln 2+\frac{1}{2}\int_0^1 \frac{\ln^2 t}{1-t}\,dt+\ln 2\left(\int_0^1 \frac{\ln t}{1-t}\,dt-\frac{1}{2}\int_0^1 \frac{\ln t}{1-t}\,dt\right)\\ &=-\frac{3}{2}\zeta(2)\ln 2+\frac{1}{2}\times 2\zeta(3)\\ &=\boxed{-\frac{3}{2}\zeta(2)\ln 2+\zeta(3)} \end{align} NB: i assume,\begin{align}R(1)&=\int_0^1 \frac{\ln t}{1-t}\,dt\\ &=-\zeta(2)\\ \int_0^1 \frac{\ln^2 t}{1-t}\,dt&=2\zeta(3)\end{align}

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  • $\begingroup$ nice approach.. $\endgroup$ – Ali Shather Sep 15 '19 at 23:43
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Start off with the substitution $x\to \frac{1-x}{1+x}$ to get: $$\require{cancel} I=\int_0^1 \frac{\ln x\ln(1+x)}{1-x}dx=\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{2}{1+x}\right)}{x}dx-\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{2}{1+x}\right)}{1+x}dx$$

$$X=\int_0^1 \frac{\ln(1-x)\ln 2 -\ln(1-x)\ln(1+x)-\ln(1+x)\ln 2+\ln^2(1+x)}{x}dx$$ $$Y=\int_0^1 \frac{\ln(1-x)\ln 2 -\ln(1-x)\ln(1+x)-\ln(1+x)\ln 2+\ln^2(1+x)}{1+x}dx$$


$$I_1=\ln 2\int_0^1 \frac{\ln(1-x)}{x}dx=\color{blue}{-\ln 2 \zeta(2)}$$ $$I_2=-\int_0^1 \frac{\ln(1-x)\ln(1+x)}{x}dx=\color{red}{\frac{5}{8}\zeta(3)}$$ $$I_3=-\ln 2 \int_0^1 \frac{\ln(1+x)}{x}dx=\color{blue}{-\frac{\ln 2}{2}\zeta(2)}$$ $$I_4=\int_0^1 \frac{\ln^2(1+x)}{x}dx=\color{red}{\frac{\zeta(3)}{4}}$$ $$I_5=\ln 2\int_0^1 \frac{\ln(1-x)}{1+x}dx=\cancel{\frac{\ln^3 2}{2}}-\cancel{\ln 2\frac{\zeta(2)}{2}}$$ $$I_6=-\int_0^1 \frac{\ln(1-x)\ln(1+x)}{1+x}dx =\cancel{-\frac{\ln^3 2}{3}}+\cancel{\ln 2\frac{\zeta(2)}{2}}-\color{red}{\frac{\zeta(3)}{8}}$$ $$I_7=-\ln 2 \int_0^1 \frac{\ln(1+x)}{1+x}dx=\cancel{-\frac{\ln^3 2}{2}}$$ $$I_8=\int_0^1 \frac{\ln^2(1+x)}{1+x}dx=\cancel{\frac{\ln^3 2}{3}}$$


$$I=X-Y=(I_1+I_2+I_3+I_4)-(I_5+I_6+I_7+I_8)=\boxed{\zeta(3)-\frac32 \ln 2 \zeta(2)}$$

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    $\begingroup$ (+1) For the brilliance of the solution. $\endgroup$ – user97357329 Sep 8 '19 at 17:08
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    $\begingroup$ nice idea Zacky. $\endgroup$ – Ali Shather Sep 8 '19 at 17:08
  • $\begingroup$ @Zacky would you mind writing the website you sent me before where i can find $\int_0^1\frac{\ln^3(1+x)\ln(1-x)}{x}\ dx$? because I posted solution there and I need it as a ref. $\endgroup$ – Ali Shather Sep 8 '19 at 19:19
  • $\begingroup$ @AliShather artofproblemsolving.com/community/u380742h1883703p12826640 $\endgroup$ – Zacky Sep 8 '19 at 20:43
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    $\begingroup$ @Zacky thank you so much $\endgroup$ – Ali Shather Sep 8 '19 at 20:48
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You might need the first generalization from the preprint A simple strategy of calculating two alternating harmonic series generalizations by Cornel Ioan Valean. All the calculations are accomplished by avoiding the evaluation of Euler sums.

$$ \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(m)}}{n}=\frac{(-1)^m}{(m-1)!}\int_0^1\frac{\displaystyle \log^{m-1}(x)\log\left(\frac{1+x}{2}\right)}{1-x}\textrm{d}x $$ $$=\frac{1}{2}\biggr(m\zeta (m+1)-2\log (2) \left(1-2^{1-m}\right) \zeta (m)$$ $$-\sum_{k=1}^{m-2} \left(1-2^{-k}\right)\left(1-2^{1+k-m}\right)\zeta (k+1)\zeta (m-k)\biggr).$$

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  • $\begingroup$ I am familiar with these generalizations .. very useful but how about to do it on our own? $\endgroup$ – Ali Shather Sep 8 '19 at 15:59
  • $\begingroup$ I started from the assumption you weren't aware of it since you didn't mention it, and then you didn't ask for ways which didn't appear in the mathematical literature yet. $\endgroup$ – user97357329 Sep 8 '19 at 16:03
  • $\begingroup$ I am very familiar with Cornel's papers but I am really into independent solutions ( if possible) that does not reply much on other results/ generalizations. $\endgroup$ – Ali Shather Sep 8 '19 at 16:22
  • $\begingroup$ Of course you are into independent solutions. The manipulations with double integrals seems to make it very easy. $\endgroup$ – user97357329 Sep 8 '19 at 16:51
  • $\begingroup$ that's the way to go :) $\endgroup$ – Ali Shather Sep 8 '19 at 16:55
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Consider the integral $$K=\int_0^1\frac{\operatorname{Li}_2(x)}{1+x}\ dx$$

By applying IBP we have

$$K=\ln(2)\zeta(2)+\int_0^1\frac{\ln(1-x)\ln(1+x)}{x}\ dx\tag{1}$$

.


On the other hand

\begin{align} K&=\int_0^1\frac{\operatorname{Li}_2(x)}{1+x}\ dx=\int_0^1\frac{1}{1+x}\left(\int_0^1-\frac{x\ln u}{1-xu}\ du\right)\ dx\\ &=\int_0^1\ln u\left(\int_0^1\frac{-x}{(1+x)(1-ux)}\ dx\right)\ du\\ &=\int_0^1\ln u\left(\frac{\ln2}{1+u}+\frac{\ln(1-u)}{u}-\frac{\ln(1-u)}{1+u}\right)\ du\\ &=\ln2\int_0^1\frac{\ln u}{1+u}\ du+\int_0^1\frac{\ln u\ln(1-u)}{u}\ du-\color{blue}{\int_0^1\frac{\ln u\ln(1-u)}{1+u}\ du}\\ &\overset{\color{blue}{IBP}}{=}\ln2\left(-\frac12\zeta(2)\right)+\zeta(3)\color{blue}{-\int_0^1\frac{\ln u\ln(1+u)}{1-u}du+\int_0^1\frac{\ln(1-u)\ln(1+u)}{u}du}\tag{2} \end{align}


Combining (1) and (2) we get

$$\int_0^1\frac{\ln x\ln(1+x)}{1-x}\ dx=\zeta(3)-\frac32\ln2\zeta(2)$$

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  • $\begingroup$ (+1) Definitely one of the best solutions. $\endgroup$ – user97357329 Sep 8 '19 at 17:02
  • $\begingroup$ Thank you @user97357329 . $\endgroup$ – Ali Shather Sep 8 '19 at 17:21
  • $\begingroup$ I think one can make things a little more simpler because $(a+b)^2-(a-b)^2=4ab$ $\endgroup$ – FDP Sep 9 '19 at 17:12
  • $\begingroup$ @FDP this method is done by Zacky here math.stackexchange.com/q/3085715 but thats not enough to solve the problem. $\endgroup$ – Ali Shather Sep 9 '19 at 17:43
  • $\begingroup$ @Ali Shather: if you want to compute $\displaystyle \int_0^1\frac{\ln(1-x)\ln(1+x)}{x}\ dx$, for me, it's enough. $\endgroup$ – FDP Sep 9 '19 at 18:54
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x} \over 1-x}\,\dd x = \zeta\pars{3} - {3 \over 2}\,\ln\pars{2}\zeta\pars{2}}:\ \approx 0.5082\ {\Large ?}}$.


\begin{align} &\bbox[5px,#ffd]{\int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x} \over 1 - x}\,\dd x} = {1 \over 2}\int_{0}^{1}{2\ln\pars{x}\ln\pars{1 + x} \over 1 - x}\,\dd x \\[5mm] = &\ {1 \over 2}\int_{0}^{1}{\ln^{2}\pars{x} + \ln^{2}\pars{1 + x} - \bracks{\ln\pars{x} - \ln\pars{1 + x}}^{\, 2} \over 1 - x}\,\dd x \\[5mm] = &\ {1 \over 2}\int_{0}^{1}{\ln^{2}\pars{x} \over 1 - x}\,\dd x + {1 \over 2}\int_{0}^{1}{\ln^{2}\pars{1 + x} - \ln^{2}\pars{2} \over 1 - x}\,\dd x \\[2mm] & \!\!\!\! -{1 \over 2}\int_{0}^{1} \bracks{\ln^{2}\pars{x \over x + 1} - \ln^{2}\pars{2}} \,{\dd x \over 1 - x} \end{align} In the $\underline{second}$ integral I'll make the change $\ds{1 + x \mapsto 2x}$ while $\ds{x/\pars{x + 1} \mapsto x}$ in the $\underline{third}$ one. Then, \begin{align} &\bbox[5px,#ffd]{\int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x} \over 1 - x}\,\dd x} \\[5mm] = &\ {1 \over 2}\int_{0}^{1}{\ln^{2}\pars{x} \over 1 - x}\,\dd x + {1 \over 2}\int_{1/2}^{1}{\ln^{2}\pars{2x} - \ln^{2}\pars{2} \over 1 - x}\,\dd x \\[2mm] & \!\!\!\! -{1 \over 2}\bracks{% -\int_{0}^{1/2}{\ln^{2}\pars{x} - \ln^{2}\pars{2} \over 1 - x} \,\dd x + 2\int_{0}^{1/2}{\ln^{2}\pars{x} - \ln^{2}\pars{2} \over 1 - 2x} \,\dd x} \\[5mm] = &\ {1 \over 2}\int_{0}^{1}{\ln^{2}\pars{x} \over 1 - x}\,\dd x + {1 \over 2}\int_{1/2}^{1}{\ln^{2}\pars{2x} - \ln^{2}\pars{2} \over 1 - x}\,\dd x \\[2mm] & \!\!\!\! + \bracks{% {1 \over 2}\int_{0}^{1/2}{\ln^{2}\pars{x} \over 1 - x}\,\dd x - {1 \over 2}\ln^{3}\pars{2} - {1 \over 2}\int_{0}^{1}{\ln^{2}\pars{x/2} - \ln^{2}\pars{2} \over 1 - x}\,\dd x} \\[5mm] & \stackrel{\mrm{IBP}}{=}\,\,\, -\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln\pars{x}\,\dd x + \bracks{% {1 \over 2}\ln^{3}\pars{2} -\int_{1/2}^{1}\mrm{Li}_{2}'\pars{x}\ln\pars{2x}\,\dd x} \\[2mm] & \phantom{\stackrel{\mrm{IBP}}{=}\,\,\,\,\,} + \bracks{{1 \over 2}\ln^{3}\pars{2} -\int_{0}^{1/2}\mrm{Li}_{2}'\pars{x}\ln\pars{x}\,\dd x} \\[2mm]& \phantom{\stackrel{\mrm{IBP}}{=}\,\,\,\,\,} -{1 \over 2}\ln^{3}\pars{2} + \int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln\pars{x \over 2}\,\dd x \\[5mm] & \stackrel{\mrm{IBP}}{=}\,\,\, \int_{0}^{1}\mrm{Li}_{3}'\pars{x}\,\dd x + {1 \over 2}\ln^{3}\pars{2} -\mrm{Li}_{2}\pars{1}\ln\pars{2} + \int_{1/2}^{1}\mrm{Li}_{3}'\pars{x}\,\dd x \\[2mm] & \phantom{\stackrel{\mrm{IBP}}{=}\,\,\,\,\,} -\mrm{Li}_{2}\pars{1 \over 2}\ln\pars{1 \over 2} + \int_{0}^{1/2}\mrm{Li}_{3}'\pars{x}\,\dd x +\mrm{Li}_{2}\pars{1}\ln\pars{1 \over 2} \\[2mm] & \phantom{\stackrel{\mrm{IBP}}{=}\,\,\,\,\,} -\int_{0}^{1}\mrm{Li}_{3}'\pars{x}\,\dd x \\[5mm] & \stackrel{\mrm{IBP}}{=}\,\,\, \zeta\pars{3} + {1 \over 2}\ln^{3}\pars{2} -2\mrm{Li}_{2}\pars{1}\ln\pars{2} + \zeta\pars{3} - \mrm{Li}_{3}\pars{1 \over 2} \\[2mm] & \phantom{\stackrel{\mrm{IBP}}{=}\,\,\,\,\,} + \mrm{Li}_{2}\pars{1 \over 2}\ln\pars{2} + \mrm{Li}_{3}\pars{1 \over 2} - \zeta\pars{3} \\[5mm] & \stackrel{\mrm{IBP}}{=}\,\,\, {1 \over 2}\ln^{3}\pars{2} -2\,\mrm{Li}_{2}\pars{1}\ln\pars{2} + \zeta\pars{3} + \mrm{Li}_{2}\pars{1 \over 2}\ln\pars{2} \label{1}\tag{1} \\[5mm] & \phantom{\stackrel{\mrm{IBP}}{=}\,\,\,\,\,} = \bbx{\large\zeta\pars{3} - {3 \over 2}\,\ln\pars{2}\zeta\pars{2}} \approx 0.5082\label{2}\tag{2} \end{align}

In (\ref{1}) and (\ref{2}) I used $\ds{\mrm{Li}_{2}\pars{1} = \zeta\pars{2} = \pi^{2}/6}$ and the known values of $\ds{\mrm{Li}_{2}\pars{1/2}}$ and $\ds{\mrm{Li}_{3}\pars{1/2}}$.

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