Help with $-\int_0^1 \ln(1+x)\ln(1-x)dx$

I have been attempting to evaluate this integral and by using wolfram alpha I know that the value is$$I=-\int_0^1 \ln(1+x)\ln(1-x)dx=\frac{\pi^2}{6}+2\ln(2)-\ln^2(2)-2$$

My Attempt:

I start off by parametizing the integral as $$I(a)=\int_0^1 -\ln(1+x)\ln(1-ax)dx$$ where $I=I(1)$. I then differentiate to get $$I'(a)=\int_0^1 \frac{ax\ln(1+x)}{1-ax}dx=\int_0^1 ax\ln(1+x)\sum_{n=0}^\infty(ax)^ndx=\sum_{n=1}^\infty a^{n+1}\int_0^1 x^{n+1}\ln(1+x)dx$$

Evaluating this integral by integration by parts and geometric series I get $$\int_0^1 x^{n+1}\ln(1+x)dx=\frac{x^{n+2}}{n+2}\ln(1+x)|_0^1-\frac{1}{n+2}\int_0^1 \frac{x^{n+2}}{1+x}dx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\int_0^1 x^{n+2}\sum_{k=0}^\infty(-x)^kdx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\sum_{k=0}^\infty(-1)^k\int_0^1 x^{k+n+2}dx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\sum_{k=0}^\infty\frac{(-1)^k}{k+n+2}=\frac{\ln(2)}{n+2}-\frac{1}{2(n+2)}\left(\psi_0\left(\frac{n}{2}+2\right)-\psi_0\left(\frac{n}{2}+\frac{3}{2}\right)\right)$$ So I arrive at $$I'(a)=\sum_{n=0}^\infty a^{n+1}\left(\frac{\ln(2)}{n+2}-\frac{1}{2(n+2)}\left(\psi_0\left(\frac{n}{2}+2\right)-\psi_0\left(\frac{n}{2}+\frac{3}{2}\right)\right)\right)$$ Re-indexing I get $$I'(a)=\frac{\ln(2)}{a}\sum_{n=2}^\infty \frac{a^n}{n}+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n}a^{n-1}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n}a^{n-1}$$Integrating both sides from $0$ to $1$ I recover $I(1)$ $$I(1)=\int_0^1 \frac{\ln(2)}{a}\left(-\ln(1-a)-a\right)da+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$ Then using the integral equation for the Dilogarithm I arrive at $$I(1)=\ln(2)\int_0^1 -\frac{\ln(1-a)}{a}da-\ln(2)+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$ $$I(1)=\frac{\ln(2)\pi^2}{6}-\ln(2)+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$

At this point I could not continue further as I did not know how to simplify the Digamma terms in the sums. I think that by using the Digamma function's relation to the Harmonic Numbers it could be possible to exploit known values of Harmonic sums to arrive at the answer but I could not get the sums in a form where this would work. If anyone could help me continue further or let me know if I am on the right track I would greatly appreciate it. Thank you in advance.

Solution 1:

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

$\ds{I \equiv -\int_{0}^{1}\ln\pars{1 + x}\ln\pars{1 - x}\,\dd x = {\pi^{2} \over 6} + 2\ln\pars{2} - \ln^{2}\pars{2} - 2:\ {\LARGE ?}}$.

\begin{align} I & \equiv \bbox[10px,#ffd]{-\int_{0}^{1}\ln\pars{1 + x}\ln\pars{1 - x}\,\dd x} \\[5mm] & = \int_{0}^{1}{\bracks{\ln\pars{1 - x} - \ln\pars{1 + x}}^{\, 2} - \bracks{\ln\pars{1 - x} + \ln\pars{1 + x}}^{\, 2} \over 4}\,\dd x \\[5mm] & = \underbrace{{1 \over 4}\int_{0}^{1}\ln^{2}\pars{1 - x \over 1 + x}\,\dd x}_{\ds{\equiv\ \mc{I}_{1}}}\ -\ \underbrace{{1 \over 4}\int_{0}^{1}\ln^{2}\pars{1 - x^{2}}\,\dd x} _{\ds{\equiv\ \mc{I}_{2}}}\ =\ \mc{I}_{1} - \mc{I}_{2}\label{1}\tag{1} \end{align}

$\ds{\Large\mc{I}_{1}:\ ?}$

\begin{align} \mc{I}_{1} & \equiv {1 \over 4}\int_{0}^{1}\ln^{2}\pars{1 - x \over 1 + x}\,\dd x \,\,\,\stackrel{\pars{1 - x}/\pars{1 + x}\ =\ t}{=}\,\,\, {1 \over 2}\int_{0}^{1}{\ln^{2}\pars{t} \over \pars{1 + t}^{2}} \,\dd t \\[5mm] & = {1 \over 2}\sum_{n = 0}^{\infty}\underbrace{-2 \choose n} _{\ds{\pars{n + 1}\pars{-1}^{n}}}\ \underbrace{\int_{0}^{1}\ln^{2}\pars{t}t^{n}\,\dd t} _{\ds{2 \over \pars{n +1}^{3}}}\ =\ \sum_{n = 1}^{\infty}{\pars{-1}^{n + 1}\over n^{2}} = \bbx{\pi^{2} \over 12}\label{2}\tag{2} \end{align}

$\ds{\Large\mc{I}_{2}:\ ?}$

\begin{align} \mc{I}_{2} & \equiv {1 \over 4}\int_{0}^{1}\ln^{2}\pars{1 - x^{2}}\,\dd x \,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\, {1 \over 8}\int_{0}^{1}x^{-1/2}\ln^{2}\pars{1 - x}\,\dd x \\[5mm] & = \left.{1 \over 8}\,\partiald[2]{}{\mu}\int_{0}^{1}x^{-1/2} \pars{1 - x}^{\mu}\,\dd x\,\right\vert_{\ \mu\ =\ 0} = {1 \over 8}\,\partiald[2]{}{\mu}\bracks{\Gamma\pars{1/2}\Gamma\pars{\mu + 1} \over \Gamma\pars{\mu + 3/2}}_{\ \mu\ =\ 0} \\[5mm] & = \bbx{-\,{\pi^{2} \over 12} - 2\ln\pars{2} + \ln^{2}\pars{2} + 2} \label{3}\tag{3} \end{align}

\eqref{1}, \eqref{2} and \eqref{3} yield the coveted result $\ds{\bbx{{\pi^{2} \over 6} + 2\ln\pars{2} - \ln^{2}\pars{2} - 2}}$.

Solution 2:

Another solution, this time using Mike Spivey's alternating Euler sum:

\begin{align*} I &=-\int_0^1\ln(1-x)\ln(1+x)\,\mathrm dx \\ &= \int_0^1\ln(1-x)\sum_{n=1}^\infty\frac{(-1)^n}{n}x^n\,\mathrm dx,\qquad\text{Mercator series}\\ &= \sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^1\ln(1-x)x^n\,\mathrm dx\\ &= \sum_{n=1}^\infty\frac{(-1)^n}{n}\cdot\frac{\partial}{\partial\alpha}\left[ \int_0^1(1-x)^\alpha x^n\,\mathrm dx\right ]\Bigg\vert_{\alpha=0}\\ &= \sum_{n=1}^\infty\frac{(-1)^n}{n}\cdot\frac{\partial}{\partial\alpha} \text{B}\left(\alpha+1,\,n+1 \right )\Bigg\vert_{\alpha=0},\qquad\text{beta function}\\ &= \sum_{n=1}^\infty\frac{(-1)^n}{n}\cdot\left[ \frac{\Gamma(\alpha+1)\Gamma(n+1)\left(\psi(\alpha+1)-\psi(\alpha+n+2) \right )}{\Gamma(\alpha+n+2)}\right ]\Bigg\vert_{\alpha=0}\\ &= -\sum_{n=1}^\infty\frac{(-1)^n}{n}\cdot\frac{H_{n+1}}{n+1},\qquad\text{used }\psi(1)=-\gamma\text{ and }\psi(m)=H_{m-1}-\gamma\\ &= -\sum_{n=1}^\infty\frac{(-1)^nH_{n+1}}{n}+\sum_{n=1}^\infty\frac{(-1)^nH_{n+1}}{n+1}\\ &= -\sum_{n=1}^\infty\frac{(-1)^nH_n}{n}-\sum_{n=1}^\infty\frac{(-1)^n}{n(n+1)}+\sum_{n=1}^\infty\frac{(-1)^nH_{n+1}}{n+1}\\ &= -\sum_{n=1}^\infty\frac{(-1)^nH_n}{n}-1+2\ln(2)+\sum_{n=1}^\infty\frac{(-1)^nH_{n+1}}{n+1}\\ &= -2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n}-2+2\ln(2)\\ &= \frac{\pi^2}{6}+2\ln(2)-\ln^2(2)-2,\qquad\text{applied Mike's sum.} \end{align*}