Evaluate $ \int_{0}^{1} \log\left(\frac{x^2-2x-4}{x^2+2x-4}\right) \frac{\mathrm{d}x}{\sqrt{1-x^2}} $

Evaluate :

$$ \int_{0}^{1} \log\left(\dfrac{x^2-2x-4}{x^2+2x-4}\right) \dfrac{\mathrm{d}x}{\sqrt{1-x^2}} $$

Introduction : I have a friend on another math platform who proposed a summation question and he has a good reputation of posting legitimate questions. I worked it out to another equivalent form i.e, the above integral. Here's my work :-

We start with $\displaystyle \sum_{n=0}^\infty L_{2n+1} x^n = \dfrac{x+1}{x^2-3x + 1} $. Replacing $x$ with $x^2$ , we get

$$ \sum_{n=0}^\infty L_{2n+1} x^{2n} = \dfrac{x^2+1}{x^4-3x^2+1} $$

Integrate:

$$ \sum_{n=0}^\infty \dfrac{L_{2n+1} x^{2n+1}}{2n+1} = \underbrace{\int \dfrac{x^2+1}{x^4-3x^2+1} \,\mathrm{d}x}_{:= I} $$

Then, $\displaystyle I = \int \dfrac{ 1+(1/x^2)}{x^2 + 1/x^2 - 3} \, \mathrm{d}x $. Let $t = x - \dfrac1x \Rightarrow \left( 1 + \dfrac1{x^2} \right) \, \mathrm{d}x = \mathrm{d}t $.

Then $\displaystyle I = \int \dfrac{\mathrm{d}t}{t^2-1} = \dfrac12 \log \left | \dfrac{t-1}{t+1} \right | = \dfrac12 \log \left | \dfrac{x^2-x-1}{x^2+x-1} \right | $.

$$ \begin{eqnarray} S & := & \sum_{n=0}^\infty \dfrac{ L_{2n+1}}{(2n+1)^2 \binom{2n}n } = \int_0^1 \sum_{n=0}^\infty \dfrac{ L_{2n+1}}{2n+1} (x-x^2)^n \, \mathrm{d}x \qquad \left(\text{ Because }\dfrac1{(2n+1) \binom{2n}n} = \operatorname{B}(n+1,n+1) = \int_{0}^{1} x^n(1-x)^n \mathrm{d}x\right) \\ &=& \int_0^1 \dfrac1{2\sqrt{x-x^2} } \log \left | \dfrac{x -x^2 - \sqrt{x-x^2} - 1}{x -x^2 + \sqrt{x-x^2} - 1} \right | \, \mathrm{d}x \\ &=& \int_0^1 f(x)\, \mathrm{d}x \end{eqnarray} $$

Note that $f(1-x) = f(x) $, so $\displaystyle S =2 \int_0^{1/2} f(x) \, \mathrm{d}x = 2 \int_0^{1/2} f\left( \dfrac12 - x\right) \, \mathrm{d}x $, and so

$$ S = \int_0^{1/2} \dfrac1{\sqrt{\frac14 - x^2}} \log \left |\dfrac{a^2-a-1}{a^2+a-1} \right| \, \mathrm{d}x $$

where $a = \sqrt{\dfrac14 - x^2} $.

Substitute $x = \dfrac12 \cos (\theta) $ and simplify:

$$ S = \int_0^{\pi/2} \log \left | \dfrac{ \cos^2 -2 \cos \theta - 4}{\cos^2 + 2\cos\theta - 4} \right | \, \mathrm{d}x = \int_0^1 \log \left ( \dfrac{x^2-2x-4}{x^2+2x-4} \right) \dfrac{\mathrm{d}x}{\sqrt{1-x^2}} \\ \vdots $$

Closed Form : Recently, the same question was posted on M.S.E. albeit in a different form by another friend of mine here. That integral is obtained from this by applying Integration By Parts. Mr. Jack D'Aurizio also gave a closed form in terms of Imaginary part of Dilogarithms, specifically, $$I = -2 \ \Im \left[\text{Li}_2\left[i\left(1-\sqrt{5}+\sqrt{5-2 \sqrt{5}}\right)\right]+\text{Li}_2\left[i\left(1+\sqrt{5}-\sqrt{5+2 \sqrt{5}}\right)\right] \right]$$ However, there is a more elementary closed form that exists for the question (as evident from the original question) in terms of natural logarithm and Catalan's constant.

All solutions are greatly appreciated.

It is really strange how often many problems here on MSE boil down to the same one.

In particular, I am talking about an identity for the squared arcsine function whose consequence is:

$$ \sum_{n\geq 0}\frac{x^n}{(2n+1)\binom{2n}{n}}=\frac{4}{\sqrt{x(4-x)}}\arcsin\left(\frac{\sqrt{x}}{2}\right)\, \tag{1}$$

If we take the following series definition of $T$: $$ T = \sum_{n\geq 0}\frac{L_{2n+1}}{(2n+1)\binom{2n}{n}}\tag{2} $$ and recall that $L_{2k+1}=\varphi^{2k+1}+\overline{\varphi}^{2k+1}$, we just have to plug in $x=\varphi^2$ and $x=\overline{\varphi}^2$ in $(1)$ to get the closed form:

$$ T = \color{red}{\frac{3\pi}{5}\sqrt{2+\frac{2}{\sqrt{5}}}-\frac{\pi}{5}\sqrt{2-\frac{2}{\sqrt{5}}}}.\tag{3}$$

If in $(1)$ we replace $x$ with $x^2 z^2$ and integrate over $[0,1]$ with respect to $x$, we get:

$$ \sum_{n\geq 0}\frac{z^{2n+1}}{(2n+1)^2 \binom{2n}{n}} = 2\int_{0}^{z/2}\frac{\arcsin(x)}{x\sqrt{1-x^2}}\,dx=2\int_{0}^{\arcsin(z/2)}\frac{\theta}{\sin\theta}\,d\theta \tag{4}$$

and now the relation with the Clausen function is self-evident.
The antiderivative of $\frac{t}{\sin t}$ is a combination of logarithms and dilogarithms and:

$$ S = 2\int_{0}^{\frac{3\pi}{10}}\frac{\theta\,d\theta}{\sin\theta}-2\int_{0}^{\frac{\pi}{10}}\frac{\theta\,d\theta}{\sin\theta}=2\int_{\pi/10}^{3\pi/10}\frac{\theta\,d\theta}{\sin\theta} \tag{5}$$

simplifies to:

$$ S = \color{red}{K+\frac{\pi}{5}\log(2)}\tag{6} $$

where: $$ K=\sum_{n\geq 0}\frac{(-1)^{n}}{(2n+1)^2}=2\int_{0}^{\pi/2}\frac{\theta\,d\theta}{\sin\theta}.\tag{7}$$

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $$ \int_{0}^{1} \ln\pars{x^{2} - 2x - 4 \over x^{2} + 2x -4}\, {\dd x \over \root{1 -x^{2}}} =\, {\Large ?} $$

$\quad$In this 'answer', I want to summarize the result which involves the Dilogarithm function $\pars{~\mbox{namely,}\ \,\mathrm{Li}_{2}\pars{z}~}$. For this purpose, I found a quite useful result in a previous Vladimir Reshetnikov answer. I'm aware of an interesting issue which has been pointed out by many users along the comments: The relationship between the '$\,\mathrm{Li}_{2}$ result' and the 'Catalan constant result' as it appears in Jack D'Aurizio answer which is somehow still open.

$$ \mathrm{f}\pars{t} = {\pi \over 2}\,\ln\pars{1 + \root{1 - t^{2}} \over 2} - 2\,\Im\,\mathrm{Li}_{2}\pars{{1 - \root{1 - t^{2}} \over t}\,\ic} $$ Note that \begin{align} &\mathrm{f}\pars{t} - \mathrm{f}\pars{-t} \\[3mm] = &\ 2\,\Im\,\mathrm{Li}_{2}\pars{-\,{1 - \root{1 - t^{2}} \over t}\,\ic} - 2\,\Im\,\mathrm{Li}_{2}\pars{{1 - \root{1 - t^{2}} \over t}\,\ic} \\[3mm] = &\ -4\,\Im\,\mathrm{Li}_{2}\pars{{1 - \root{1 - t^{2}} \over t}\,\ic} \end{align}

Since $\ds{\Phi^{2} + \Phi + 1 = \varphi^{2} - \varphi - 1 = 0}$: \begin{align} \left.{1 - \root{1 - t^{2}} \over t}\right\vert_{\ t\ =\ \Phi/2} = {2 - \root{3 + \Phi} \over \Phi}\quad\mbox{and}\quad \left.{1 - \root{1 - t^{2}} \over t}\right\vert_{\ t\ =\ \varphi/2} = {2 - \root{3 - \varphi} \over \varphi} \end{align} Then, \begin{align} &\color{#f00}{\int_{0}^{1}\ln\pars{x^{2} - 2x - 4 \over x^{2} + 2x -4} \,{\dd x \over \root{1 -x^{2}}}} \\[3mm] = &\ \color{#f00}{% 4\,\Im\,\mathrm{Li}_{2}\pars{{2 - \root{3 - \varphi} \over \varphi}\,\ic} - 4\,\Im\,\mathrm{Li}_{2}\pars{{2 - \root{3 + \Phi} \over \Phi}\,\ic}} \\[3mm] = &\ \color{#f00}{% 4\,\Im\,\mathrm{Li}_{2}\pars{\bracks{-1 + \root{5} - \root{5 - 2\root{5}}}\ic} - 4\,\Im\,\mathrm{Li}_{2}\pars{\bracks{1 + \root{5} - \root{5 + 2\root{5}}}\ic}} \tag{1} \\[3mm] \approx &\ 1.3523870463919131106825397783200\color{#f00}{513308068289818222} \end{align}

This is slightly bigger $\pars{~\sim 1.16\times 10^{-32}~}$ than the 'direct' numerical calculation of the original integral $\pars{~\approx 1.3523870463919131106825397783200\color{#f00}{397004399596207528}~}$. ADDENDA The user @Ishan Singh shows me ( $\mbox{01-jul-2016}$ ) a closed expression: \begin{align} &-\,{\pi \over 5}\, \ln\pars{124 - 55\root{5} + 2\root{7625 - 3410\root{5}}} \\[2mm] &\ + {8 \over 5}\,G \tag{2} \end{align} where $\ds{G}$ is the Catalan Constant. It would be nice to know 'how to travel' between $\pars{1}$ and $\pars{2}$.