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

Question

I have been trying to derive

$$\int_0^1\frac{\ln(1-x)\ln(x+x^2)}{1+x^2}dx = \frac{\pi^3}{64} +\frac\pi{16}\ln^22-G\ln2$$

with $G$ being the the Catalan constant.

I noticed that a similarly-looking integral is posted and solved here. Although the solution is applicable and meritorious in itself, it seems an overkill to resort to the special function $\operatorname{Li}_3(z)$ given the elementary result.

Answer 1

Incomplete solution for now. \begin{align}A&=\int_0^1\frac{\ln(1-x)\ln x}{1+x^2}dx\\ B&=\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x^2}dx\\ J&=A+B\\ A&\overset{y=\frac{1-x}{1+x}}=\int_0^1 \frac{\ln\left(\frac{2y}{1+y}\right)\ln\left(\frac{1-y}{1+y}\right)}{1+y^2}dy\\ &=\int_0^1 \frac{\ln 2\ln\left(\frac{1-u}{1+u}\right)}{1+u^2}du+\int_0^1 \frac{\ln x\ln\left(\frac{1-y}{1+y}\right)}{1+y^2}dy+\int_0^1 \frac{\ln^2\left(1+y\right)}{1+y^2}dy-B\\ &\overset{x=\frac{1-u}{1+u}}=-\text{G}\ln 2+\int_0^1 \frac{\ln x\ln\left(\frac{1-y}{1+y}\right)}{1+y^2}dy+\int_0^1 \frac{\ln^2\left(1+y\right)}{1+y^2}dy-B\\ &\overset{y=\tan x}=-\text{G}\ln 2+\int_0^{\frac{\pi}{4}}\ln\left(\tan x\right)\ln\left(\tan\left(\frac{\pi}{4}-x\right)\right)dx+\int_0^{\frac{\pi}{4}}\ln^2\left(1+\tan x\right)dx-B\\ &=-\text{G}\ln 2+\int_0^{\frac{\pi}{4}}\ln\left(\frac{\sin x}{\cos x}\right)\ln\left(\frac{\sin\left(\frac{\pi}{4}-x\right)}{\cos\left(\frac{\pi}{4}-x\right)}\right)dx+\int_0^{\frac{\pi}{4}}\ln^2\left(\frac{\sqrt{2}\cos\left(\frac{\pi}{4}-x\right)}{\cos x}\right)dx-B\\ &=-\text{G}\ln 2+\int_0^{\frac{\pi}{4}}\ln\left(\sin x\right)\ln\left(\sin\left(\frac{\pi}{4}-x\right)\right)dx-\int_0^{\frac{\pi}{4}}\ln\left(\sin u\right)\ln\left(\cos\left(\frac{\pi}{4}-u\right)\right)du-\\& \int_0^{\frac{\pi}{4}}\ln\left(\cos x\right)\ln\left(\sin\left(\frac{\pi}{4}-x\right)\right)dx+ \int_0^{\frac{\pi}{4}}\ln\left(\cos x\right)\ln\left(\cos\left(\frac{\pi}{4}-x\right)\right)dx+\int_0^{\frac{\pi}{4}}\ln^2\left(\sqrt{2}\right)dx+\\& \int_0^{\frac{\pi}{4}}\ln^2\left(\cos\left(\frac{\pi}{4}-u\right)\right)du+\underbrace{2\int_0^{\frac{\pi}{4}}\ln\left(\sqrt{2}\right)\ln\left(\cos\left(\frac{\pi}{4}-u\right)\right)du-2\int_0^{\frac{\pi}{4}}\ln\left(\sqrt{2}\right)\ln\left(\cos x\right)dx}_{=0}\\& +\int_0^{\frac{\pi}{4}}\ln^2\left(\cos x\right)dx-2\int_0^{\frac{\pi}{4}}\ln\left(\cos x\right)\ln\left(\cos\left(\frac{\pi}{4}-x\right)\right)dx-B\\ &\overset{x=\frac{\pi}{4}-u}=-\text{G}\ln 2+\int_0^{\frac{\pi}{4}}\ln\left(\sin x\right)\ln\left(\sin\left(\frac{\pi}{4}-x\right)\right)dx-2\int_0^{\frac{\pi}{4}}\ln\left(\cos x\right)\ln\left(\sin\left(\frac{\pi}{4}-x\right)\right)dx-\\&\int_0^{\frac{\pi}{4}}\ln\left(\cos x\right)\ln\left(\cos\left(\frac{\pi}{4}-x\right)\right)dx+2\int_0^{\frac{\pi}{4}}\ln^2\left(\cos x\right)dx+\frac{\pi\ln^2 2}{16}-B\\ \end{align} From Evaluating $\int_0^{\pi/4} \ln(\tan x)\ln(\cos x-\sin x)dx=\frac{G\ln 2}{2}$ one obtains: \begin{align}\int_0^{\frac{\pi}{4}}\ln\left(\sin x\right)\ln\left(\sqrt{2}\sin\left(\frac{\pi}{4}-x\right)\right)dx-\int_0^{\frac{\pi}{4}}\ln\left(\cos x\right)\ln\left(\sqrt{2}\sin\left(\frac{\pi}{4}-x\right)\right)dx&=\frac{\text{G}\ln 2}{2}\\ \frac{\ln 2}{2}\int_0^{\frac{\pi}{4}}\ln\left(\tan x\right)dx+\int_0^{\frac{\pi}{4}}\ln\left(\sin x\right)\ln\left(\sin\left(\frac{\pi}{4}-x\right)\right)dx-\int_0^{\frac{\pi}{4}}\ln\left(\cos x\right)\ln\left(\sin\left(\frac{\pi}{4}-x\right)\right)dx&=\frac{\text{G}\ln 2}{2}\\ \int_0^{\frac{\pi}{4}}\ln\left(\sin x\right)\ln\left(\sin\left(\frac{\pi}{4}-x\right)\right)dx-\int_0^{\frac{\pi}{4}}\ln\left(\cos x\right)\ln\left(\sin\left(\frac{\pi}{4}-x\right)\right)dx&=\text{G}\ln 2\\ \end{align}

Therefore, \begin{align}J&=\text{G}\ln 2-\int_0^{\frac{\pi}{4}}\ln\left(\sin x\right)\ln\left(\sin\left(\frac{\pi}{4}-x\right)\right)dx-\int_0^{\frac{\pi}{4}}\ln\left(\cos x\right)\ln\left(\cos\left(\frac{\pi}{4}-x\right)\right)dx+\\&2\int_0^{\frac{\pi}{4}}\ln^2\left(\cos x\right)dx+\frac{\pi\ln^2 2}{16}\\ &=\text{G}\ln 2-\frac{1}{2}\left(\int_0^{\frac{\pi}{4}}\ln^2\left(\sin x\right)dx+\int_0^{\frac{\pi}{4}}\ln^2\left(\sin \left(\frac{\pi}{4}-u\right)\right)du-\int_0^{\frac{\pi}{4}}\ln^2\left(\frac{\sin \left(\frac{\pi}{4}-x\right)}{\sin x}\right)dx\right)-\\& \frac{1}{2}\left(\int_0^{\frac{\pi}{4}}\ln^2\left(\cos x\right)dx+\int_0^{\frac{\pi}{4}}\ln^2\left(\cos \left(\frac{\pi}{4}-u\right)\right)du-\int_0^{\frac{\pi}{4}}\ln^2\left(\frac{\cos \left(\frac{\pi}{4}-x\right)}{\cos x}\right)dx\right)+\\ &2\int_0^{\frac{\pi}{4}}\ln^2\left(\cos x\right)dx+\frac{\pi\ln^2 2}{16}\\ &\overset{x=\frac{\pi}{4}-u}=\text{G}\ln 2-\int_0^{\frac{\pi}{4}}\ln^2\left(\sin x\right)dx+\int_0^{\frac{\pi}{4}}\ln^2\left(\cos x\right)dx+\frac{1}{2}\int_0^{\frac{\pi}{4}}\ln^2\left(\frac{\sin \left(\frac{\pi}{4}-x\right)}{\sin x}\right)dx+\\&\frac{1}{2}\int_0^{\frac{\pi}{4}}\ln^2\left(\frac{\cos \left(\frac{\pi}{4}-x\right)}{\cos x}\right)dx+\frac{\pi\ln^2 2}{16}\\ \end{align} I think the roadmap is clear. I will terminate later, i hope, . Busy for now.

Addendum number 1, one step beyond !

\begin{align}\int_0^{\frac{\pi}{4}}\ln^2\left(\sin x\cos x\right)dx+\underbrace{\int_0^{\frac{\pi}{4}}\ln^2\left(\tan x\right)dx}_{=\frac{\pi^3}{16}}&=2\int_0^{\frac{\pi}{4}}\ln^2\left(\sin x\right)dx+2\int_0^{\frac{\pi}{4}}\ln^2\left(\cos x\right)dx\\ \int_0^{\frac{\pi}{4}}\ln^2\left(\sin x\cos x\right)dx&=\int_0^{\frac{\pi}{4}}\ln^2\left(\frac{\sin(2x)}{2}\right)dx\\ &\overset{y=2x}=\frac{1}{2}\int_0^{\frac{\pi}{2}}\ln^2\left(\frac{\sin y}{2}\right)dy\\ &=\frac{1}{2}\int_0^{\frac{\pi}{2}}\ln^2\left(\sin y\right)dy+\frac{\pi\ln^2 2}{4}-\\&\ln 2\int_0^{\frac{\pi}{2}}\ln\left(\sin x\right)dx\\ &=\frac{1}{2}\int_0^{\frac{\pi}{2}}\ln^2\left(\sin y\right)dy+\frac{3\pi\ln^2 2}{4} \end{align} Moreover, \begin{align}\underbrace{\int_0^{\frac{\pi}{2}}\ln^2\left(\tan x\right)dy}_{=\frac{\pi^3}{8}}+\int_0^{\frac{\pi}{2}}\ln^2\left(\sin x\cos x\right)dy&=\\2\int_0^{\frac{\pi}{2}}\ln^2\left(\sin y\right)dy+2\underbrace{\int_0^{\frac{\pi}{2}}\ln^2\left(\cos u\right)du}_{y=\frac{\pi}{2}-u}&\\ \frac{\pi^3}{8}+\underbrace{\int_0^{\frac{\pi}{2}}\ln^2\left(\frac{\sin(2x)}{2}\right)dx}_{y=2x}&=4\int_0^{\frac{\pi}{2}}\ln^2\left(\sin y\right)dy\\ \frac{\pi^3}{8}+\frac{1}{2}\int_0^{\frac{\pi}{2}}\ln^2\left(\frac{\sin y}{2}\right)dy&=4\int_0^{\frac{\pi}{2}}\ln^2\left(\sin y\right)dy\\ \frac{\pi^3}{8}+\frac{1}{2}\int_0^{\pi}\ln^2\left(\sin y\right)dy+\frac{\pi\ln^2 2}{2}-\ln 2\int_0^{\pi}\ln\left(\sin y\right)dy&=4\int_0^{\frac{\pi}{2}}\ln^2\left(\sin y\right)dy\\ \frac{\pi^3}{8}+\frac{1}{2}\left(\int_0^{\frac{\pi}{2}}\ln^2\left(\sin y\right)dy+\underbrace{\int_{\frac{\pi}{2}}^\pi\ln^2\left(\sin u\right)du}_{x=\pi-u}\right)-\\ \ln 2\left(\int_0^{\frac{\pi}{2}}\ln\left(\sin y\right)dy+\underbrace{\int_{\frac{\pi}{2}}^\pi\ln\left(\sin u\right)du}_{x=\pi-u}\right)+\frac{\pi\ln^2 2}{2}=4\int_0^{\frac{\pi}{2}}\ln^2\left(\sin y\right)dy\\ \end{align} Therefore, \begin{align}\int_0^{\frac{\pi}{2}}\ln^2\left(\sin y\right)dy&=\frac{\pi^3}{24}+\frac{\pi\ln^2 2}{2}\\ \int_0^{\frac{\pi}{4}}\ln^2\left(\sin x\cos x\right)dx&=\frac{\pi^3}{48}+\pi\ln^2 2\\ \int_0^{\frac{\pi}{4}}\ln^2\left(\sin x\right)dx&=\frac{\pi^3}{24}+\frac{\pi\ln^2 2}{2}-\int_0^{\frac{\pi}{4}}\ln^2\left(\cos x\right)dx\\ J&=\text{G}\ln 2-\frac{\pi^3}{24}-\frac{7\pi\ln^2 2}{16}+2\int_0^{\frac{\pi}{4}}\ln^2\left(\cos x\right)dx+\\&\frac{1}{2}\int_0^{\frac{\pi}{4}}\ln^2\left(\frac{\sin \left(\frac{\pi}{4}-x\right)}{\sin x}\right)dx+\frac{1}{2}\int_0^{\frac{\pi}{4}}\ln^2\left(\frac{\cos \left(\frac{\pi}{4}-x\right)}{\cos x}\right)dx\\ \end{align} (NB:the second integral is doable, a closed-form is $\dfrac{5\pi^3}{64}$, not sure the last one is easily doable.)

To be continued...

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{0}^{1}{\ln\pars{1 - x}\ln\pars{x + x^{2}} \over 1 + x^{2}}\,\dd x} \\[5mm] = &\ {\pi^{3} \over 64} + {\pi \over 16}\ln^{2}\pars{2} -G\ln\pars{2} \\[2mm] \approx &\ -0.0560891672816387\ldots:\ {\Large ?}. \end{align}

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$\ds{\Large\overline{\underline{\mbox{Work in Progress !!!}}}:}$ \begin{align} &\bbox[5px,#ffd]{\int_{0}^{1}{\ln\pars{1 - x} \ln\pars{x + x^{2}} \over 1 + x^{2}}\,\dd x} \\[5mm] = &\ -\,\Im\int_{0}^{1}{\ln\pars{1 - x} \ln\pars{x + x^{2}} \over \ic - x}\,\dd x \\[5mm] = &\ -\underbrace{\Im\int_{0}^{1}{\ln\pars{1 - x} \ln\pars{x} \over \ic - x}\,\dd x} _{\ds{-{\cal J}_{1}}} \\[2mm] - &\ \underbrace{\Im\int_{0}^{1}{\ln\pars{1 - x} \ln\pars{1 + x} \over \ic - x}\,\dd x} _{\ds{-{\cal J}_{2}}}\ =\ {\cal J}_{1} + {\cal J}_{2}\label{1}\tag{1} \end{align}

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$\ds{\large{\cal J}_{1}\ \mbox{Evaluation:}}$ \begin{align} {\cal J}_{1} & \equiv -\,\Im\int_{0}^{1}{\ln\pars{1 - x} \ln\pars{x} \over \ic - x}\,\dd x \\[5mm] = &\ -\,{1 \over 2}\Im\int_{0}^{1}{\ln^{2}\pars{1 - x} \over \ic - x}\,\dd x -\,{1 \over 2}\Im\int_{0}^{1}{\ln^{2}\pars{x} \over \ic - x}\,\dd x \\[2mm] + &\ {1 \over 2}\Im\int_{0}^{1}\ln^{2}\pars{1 - x \over x}\,{\dd x \over \ic - x} \end{align} I'll make the changes $\ds{\pars{~x \mapsto 1 - x~}}$ and $\ds{\pars{~{1 - x \over x} \mapsto x~}}$ in the first and in the third integral, respectively: \begin{align} {\cal J}_{1} = &\ {1 \over 2}\Im\int_{0}^{1}{\ln^{2}\pars{x} \over 1 - \ic - x}\,\dd x -\,{1 \over 2}\Im\int_{0}^{1}{\ln^{2}\pars{x} \over \ic - x}\,\dd x \\[2mm] - &\ {1 \over 2}\Re\int_{0}^{\infty} {\ln^{2}\pars{x} \over \pars{1 + x}\bracks{\pars{1 + \ic} + x}}\,\dd x \end{align} The integrands in those expressions can be reduced or/and related to the form $\ds{\ln^{2}\pars{x} \over a - x}$ which is straightforward evaluated by integrating twice by parts: $$ -\ln\pars{1 - {x \over a}}\ln^{2}\pars{x} - 2\ln\pars{x}\on{Li}_{2}\pars{x \over a} + 2\on{Li}_{3}\pars{x \over a} $$ Namely, $$ \left\{\begin{array}{rcl} \ds{{1 \over 2}\Im\int_{0}^{1}{\ln^{2}\pars{x} \over 1 - \ic - x}\,\dd x} & \ds{=} & \ds{\Im\on{Li}_{3}\pars{1 + \ic \over 2}} \\[2mm] \ds{-\,{1 \over 2}\Im\int_{0}^{1}{\ln^{2}\pars{x} \over \ic - x}\,\dd x} & \ds{=} & \ds{\pi^{3} \over 32} \end{array}\right. $$ $\ds{\underline{\mbox{and}}}$ \begin{align} &-{1 \over 2}\Re\int_{0}^{\infty} {\ln^{2}\pars{x} \over \pars{1 + x}\bracks{\pars{1 + \ic} + x}}\,\dd x = -{5\pi^{3} \over 128} - {\pi\ln^{2}\pars{2} \over 32} \end{align} In adding those expressions, I found: \begin{align} {\cal J}_{1} & \equiv -\,\Im\int_{0}^{1}{\ln\pars{1 - x} \ln\pars{x} \over \ic - x}\,\dd x \\[5mm] = &\ -{\pi^{3} \over 128} - {\pi\ln^{2}\pars{2} \over 32} + \Im\on{Li}_{3}\pars{1 + \ic \over 2} \approx 0.2807 \end{align}

Answer 3

$$I=\int_{0}^{1}\frac{log(1-x)log(x+x^2)}{1+x^2}dx$$

$$=\underbrace{\int_{0}^{1}\frac{log(1-x)log(x)}{1+x^2}dx}_{I_1}+\underbrace{\int_{0}^{1}\frac{log(1-x)log(1+x)}{1+x^2}dx}_{I_2}$$

Let's start evaluating $I_2$ using Leibniz Rule: $$I_2={\int_0^1}\frac{log(1-x)log(1+x)}{1+x^2}dx=-{\int_0^1}{\int_0^1}{\int_0^1}\frac{x^2}{(1+x^2)(1-yx)(1+xz)}dydzdx$$

$$={\int_0^1}{\int_0^1}{\int_0^1}dxdydz\\\left(\frac{(1+yz)+x(y-z)}{(1+y^2)(1+z^2)(1+x^2)}-\frac{y}{(1+y^2)(y+z)(1-yz)}-\frac{z}{(1+z^2)(y+z)(1+zx)}\right)$$

$$={\int_0^1}{\int_0^1}dydz\left(\frac{\frac{\pi}{4}(1+yz)+\frac{log(2)}{2}(y-z)}{(1+y^2)(1+z^2)}+\underbrace{\frac{log(1-y)}{(1+y^2)(y+z)}}_{y\rightarrow z\\ z\rightarrow y}+\frac{log(1+z)}{(1+z^2)(y+z)}\right)$$

$$={\int_0^1}{\int_0^1}dydz\left(\frac{\frac{\pi}{4}(1+yz)+\frac{log(2)}{2}(y-z)}{(1+y^2)(1+z^2)}+\frac{log\left(\frac{1-z}{1+z}\right)}{(1+z^2)(y+z)}\right)$$

$$=\int_0^1\left(\frac{\frac{\pi^2}{16}+\frac{log^2(2)}{2}}{1+z^2}+\underbrace{\frac{log\left(\frac{1-z}{1+z}\right)log\left(\frac{1+z}{z}\right)}{1+z^2}}_{z\rightarrow \frac{1-x}{1+x}}\right)dz$$

$$=\frac{\pi}{4}\left(\frac{\pi^2}{16}+\frac{log^2(2)}{4}\right)+log(2)\int_0^1\frac{log(x)}{1+x^2}dx-\underbrace{\int_{0}^{1}\frac{log(x)log(1-x)}{1+x^2}dx}_{I_1}$$

$$=\frac{\pi^3}{64}+\frac{\pi}{16}log^2(2)-Glog(2)-I_1$$

Hence: $$I=I_1+I_2=\int_{0}^{1}\frac{log(1-x)log(x+x^2)}{1+x^2}dx =\frac{\pi^3}{64}+\frac{\pi}{16}log^2(2)-Glog(2)\ \blacksquare$$

Answer 4

Write the numerator of the integrand as $$\ln(1-x)\ln(x+x^2)=\ln(1-x)\ln x+\ln(1+x)\ln\frac{1-x}{1+x}+\ln^2(1+x) $$ and, correspondingly

$$I=\int_0^1\frac{\ln(1-x)\ln(x+x^2)}{1+x^2}dx = I_1+I_2+I_3 $$ where \begin{align} I_1 &= \int_0^1\frac{\ln(1-x)\ln x}{1+x^2}dx = \frac12\int_0^1\frac{\ln^2(1-x)+\ln^2x-\ln^2\frac x{1-x}}{1+x^2}dx\\ I_2 &=\int_0^1\frac{\ln(1+x)\ln \frac{1-x}{1+x}}{1+x^2}dx \overset{ \frac{1-x}{1+x} \to x} = -\int_0^1\frac{\ln x\ln(1+x)}{1+x^2}dx-G\ln2 \\ I_3 &= \int_0^1\frac{\ln^2(1+x)}{1+x^2}dx = \int_0^\infty\frac{\ln^2(1+x)}{1+x^2}dx- \int_1^\infty \frac{\ln^2(1+x)}{1+x^2} \overset{x\to 1/x} {dx }\\ &= \int_0^\infty\frac{\ln^2(1+x)}{1+x^2}dx- \left( I_3 + \int_0^1 \frac{\ln^2x}{1+x^2}dx-2 \int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}dx\right)\\ &=\frac12 \int_0^\infty\frac{\ln^2(1+x)}{1+x^2}dx - \frac12 \int_0^1\frac{\ln^2 x}{1+x^2}dx +\int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}dx \end{align} Then, add up the three integrals above to obtain \begin{align} I=& \ I_1+I_2+I_3\\ =& \ \frac12 \int_0^1\overset{t=1-x}{\frac{\ln^2(1-x)}{1+x^2}}dx +\frac12\int_0^\infty\overset{t=1+x}{ \frac{\ln^2(1+x)}{1+x^2}}dx - \frac12\int_0^1\overset{t=x/(1-x)}{ \frac{\ln^2\frac x{1-x}}{1+x^2}}dx-G\ln2 \\ =& \ 2\int_0^\infty \underset{t^2\to 2t} {\frac{t\ln^2t}{t^4+4}dt } -G\ln2 =\frac{\pi^3}{64} +\frac\pi{16}\ln^22-G\ln2 \end{align}