Evaluating $\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x\right)}{1+x^2}\:dx$

Question

I've been trying to evaluate $$\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x\right)}{1+x^2}\:dx$$ With no success, i tried to consider the following integrals $$I=\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x\right)}{1+x^2}\:dx,J=\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1+x\right)}{1+x^2}\:dx$$ $$I+J=\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x^2\right)}{1+x^2}\:dx=\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x^4\right)}{1+x^2}\:dx-\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1+x^2\right)}{1+x^2}\:dx$$ I managed to express that $1$st integral into somewhat known euler sums but that $2$nd integral arrived at a sum i didnt know how to evaluate which was $$2\sum _{k=1}^{\infty }\frac{\left(-1\right)^kH_k}{\left(2k+1\right)^3}$$ And it seems this approach wont go smooth, could i tackle the main integral differently? maybe with an easier approach?

Answer 1

Integrate elementarily as follows

\begin{align} &\int _0^1\frac{\ln^2x\ln \left(1-x\right)}{1+x^2}\:dx\\ =& \int _0^1 \ln (1-x)\>d\left( \int_1^x \frac{\ln^2t}{1+t^2}\right) = \int _0^1 \frac{ \int_0^x \frac{\ln^2t}{1+t^2}\overset{t=xy}{dt}- \int_0^1 \frac{\ln^2t}{1+t^2}dt}{1-x}dx\\ = & \int _0^1 \left(\int_0^1 \frac{\ln^2(xy)}{1+y^2}\bigg( \frac{xy^2-1}{1+x^2y^2}+\frac1{1-x}\right) dy-\frac{ \int_0^1 \frac{\ln^2t}{1+t^2}dt}{1-x}\bigg)dx\\ = & \int _0^1 \int_0^1 \frac{\ln^2x+2\ln x\ln y}{(1-x)(1+y^2)} \>dy \>dx+\int _0^1 \int_0^1 \frac{\ln^2(xy)(xy^2-1)}{(1+y^2)(1+x^2y^2)} \overset{x=\frac ty}{dx}\>dy\\ = & \>2\zeta(3)\frac\pi4 +2\left(-\frac{\pi^2}6\right)(-G)+\int _0^1 \frac1{y(1+y^2)} {\int_0^y} \frac{\ln^2t\>(yt-1)}{1+t^2}dt\> \overset{ibp}{dy}\\ =&\>\frac\pi2\zeta(3)+\frac{\pi^2}3G +\frac\pi4\int_0^1 \underset{=3\zeta(3)/16}{\frac{{t\ln^2t}}{1+t^2}}dt-\frac12\ln2 \int_0^1 \underset{=\pi^3/16}{\frac{{\ln^2t}}{1+t^2}}dt\\ & +\int_0^1 \frac{{\ln^3y}}{\underset{=-6\beta(4)} {1+y^2}}dy -\frac12 \int_0^1 \frac{\ln^2y\ln(1+y^2)}{\underset{=K_1} {1+y^2}}dy-\int_0^1 \frac{y\ln^2y\tan^{-1}y}{\underset{=K_2}{1+y^2}}dy\\ =&-6\beta(4)+\frac{35\pi}{64}\zeta(3)+\frac{\pi^2}3G+\frac{\pi^3}{32}\ln2 -\frac12K_1-K_2\hspace{2cm}(1) \end{align} Note that \begin{align} K_1= &\ \bigg(\int_0^\infty-\overset{y\to 1/y}{\int_1^\infty}\bigg) \frac{\ln^2y\ln(1+y^2)}{1+y^2}dy\\ =&\ \frac12 \int_0^\infty \frac{\ln^2y\ln(1+y^2)}{1+y^2}dy-6\beta(4)\\ K_2=& \int_0^1 \frac{\ln^2y\tan^{-1}y}y\overset{ibp}{dy}- \bigg(\int_0^\infty-\overset{y\to 1/y}{\int_1^\infty}\bigg) \frac{\ln^2y\tan^{-1}y}{y(1+y^2)}dy\\ =& \>\beta(4)+\frac{3\pi}{64}\zeta(3)-\int_0^\infty \frac{\ln^2y\tan^{-1}y}{y(1+y^2)}dy \end{align} Then, (1) becomes $$\int _0^1\frac{\ln^2x\ln \left(1-x\right)}{1+x^2}\:dx = -4\beta(4)+\frac\pi2 \zeta(3)+\frac{\pi^2}3G+\frac{\pi^3}{32}\ln2+\frac12 J(1)\tag2 $$ where $J(a)=\int_0^\infty \frac{\ln^2y\>[\frac1y\tan^{-1}(ay)-\frac12 \ln(1+a^2y^2)]}{1+y^2}dy$ and \begin{align} J’(a)=\int_0^\infty \frac{\ln^2y\>(1-ay^2)}{(1+y^2)(1+a^2y^2)}dy=-\frac{\pi\ln^2a}{2(1-a)}\\ \end{align} which leads to $J(1) =\int_0^1 J’(a)\>da = -\pi\zeta(3)$. Plug into (2) to obtain $$\int _0^1\frac{\ln^2x\ln \left(1-x\right)}{1+x^2}\:dx = -4\beta(4)+\frac{\pi^2}3G+\frac{\pi^3}{32}\ln2 $$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[10px,#ffd]{\int_{0}^{1} {\ln^{2}\pars{x}\ln\pars{1 - x} \over 1 + x^{2}}\,\dd x}:\ {\Large ?}}$.

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\begin{align} &\mbox{Lets consider} \\[1mm] &\ \mathcal{I}\pars{a} \equiv \left.\int_{0}^{1} {\ln^{2}\pars{x}\ln\pars{1 - ax} \over 1 + x^{2}}\,\dd x\, \right\vert_{\ a\ >\ 1}\,,\ \mathcal{I}\pars{0} = 0 \label{1}\tag{1} \end{align}

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\begin{align} \mathcal{I}'\pars{a} & \equiv \Im\int_{0}^{1}{x\ln^{2}\pars{x} \over \pars{\ic - x}\pars{1 - ax}}\,\dd x \\[5mm] & = -\,\Im\bracks{{1 \over a + \ic}\int_{0}^{1}{\ln^{2}\pars{x} \over \ic - x}\,\dd x} - \Im\bracks{{\ic/a \over a + \ic}\int_{0}^{1}{\ln^{2}\pars{x} \over 1/a - x}\,\dd x} \end{align} However, $\ds{\int_{0}^{1}{\ln^{2}\pars{x} \over \xi - x}\,\dd x = 2\,\mrm{Li}_{3}\pars{1 \over \xi}}$. Then, \begin{align} \mathcal{I}'\pars{a} & = -2\,\Im\bracks{{\mrm{Li}_{3}\pars{-\ic} \over a + \ic}} - 2\,\Re\bracks{\mrm{Li}_{3}\pars{a} \over a\pars{a + \ic}} \\[5mm] \mathcal{I}\pars{1} & = -2\,\Im\bracks{\mrm{Li}_{3}\pars{-\ic}\int_{0}^{1}{\dd a \over a + \ic}} + 2\,\Im\int_{0}^{1}{\mrm{Li}_{3}\pars{a} \over \ic + a}\,\dd a \end{align}

Answer 3

Using the same approach as @Felix Marin, making the story short, we have $$I'(a)=\frac{\pi ^3 a-32 \text{Li}_3(a)-3 \zeta (3)}{16( a^2+1)}$$ $$\int I'(a)\,da=\frac{\pi ^3}{32} \log \left(a^2+1\right)-\frac{3\zeta (3)}{16} \tan ^{-1}(a)-2\int\frac{ \text{Li}_3(a)}{a^2+1}\,da$$ $$\int_0^1 I'(a)\,da=\frac{\pi}{64} \left(2\pi ^2 \log (2)-3 \zeta (3)\right)-2\int_0^1\frac{ \text{Li}_3(a)}{a^2+1}\,da$$

As said in comments, the last integral is given here and then the final result.