TITLE: Evaluate $ \int_{0}^{1} \log\left(\frac{x^2-2x-4}{x^2+2x-4}\right) \frac{\mathrm{d}x}{\sqrt{1-x^2}} $
QUESTION [16 upvotes]: 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.

REPLY [6 votes]: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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$$
\int_{0}^{1}
\ln\pars{x^{2} - 2x - 4 \over x^{2} + 2x -4}\,
{\dd x \over \root{1 -x^{2}}} =\, {\Large ?}
$$
$\quad$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.

The roots of $\ds{x^{2} - 2x - 4}$ are given by $\ds{2\varphi}$ and $-2\ds{\Phi}$ where $\ds{\varphi}$ and $\ds{\Phi = 1/\varphi}$
are the *G0lden Ratio* $\ds{\root{5} + 1 \over 2}$ and the *Conjugated Golden Ratio* $\ds{\root{5} - 1 \over 2}$, respectively. Similarly, the roots of $\ds{x^{2} + 2x - 4}$ are given by $\ds{-2\varphi}$ and $\ds{2\Phi}$. Namely,
\begin{align}
&x^{2} - 2x - 4 = \pars{x - 2\varphi}
\pars{x + 2\Phi}
\\[5mm] = &\
-4\pars{1 - \half\,\Phi x}\pars{1 + \half\,\varphi x}
\\[3mm] &\
x^{2} + 2x - 4 = \pars{x + 2\varphi}\pars{x - 2\Phi}
\\[5mm] = &\
-4\pars{1 + \half\,\Phi x}\pars{1 - \half\,\varphi x}
\end{align}
With the sub$\ds{\ldots\ x = \cos\pars{\theta}}$, the above integral is rewritten as:
\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] = &\
\bracks{\mathrm{f}\pars{\half\,\Phi} - \mathrm{f}\pars{-\,\half\,\Phi}} -
\bracks{%
\mathrm{f}\pars{\half\,\varphi} - \mathrm{f}\pars{-\,\half\,\varphi}}
\end{align}
where
$$
\mathrm{f}\pars{t} \equiv 
\int_{0}^{1}{\ln\pars{1 - tx} \over \root{1 - x^{2}}}\,\dd x =
\int_{0}^{\pi/2}\ln\pars{1 - t\cos\pars{\theta}}\,\dd\theta\,,\qquad t \in \pars{-1,1}
$$
$\mathrm{f}\pars{t}$ is given by [formula $\pars{4}$ in another Vladimir Reshetnikov answer][1]:
$$
\mathrm{f}\pars{t} =
{\pi \over 2}\,\ln\pars{1 + \root{1 - t^{2}} \over 2} -
2\,\Im\,\mathrm{Li}_{2}\pars{{1 - \root{1 - t^{2}} \over t}\,\ic}
$$
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 $\pars{2}$.