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

• Thanks for the answer, but I think you missed a $(2n+1)$ in the denominator (of $S$). Also, can we directly evaluate the integral in the question? Jun 15, 2016 at 17:08
• @IshanSingh: thank you, I fixed my answer. My $(4)$ now is evidently the reason for $K$ (the Catalan constant) to appear in the closed form, as half the integral over $(0,\pi/2)$ of $\frac{\theta}{\sin\theta}$. Jun 15, 2016 at 17:20
• I think you misinterpreted my question, I was asking that whether we can show the identity (which we get if we consider both your answers) $$-2 \ \Im \left[\text{Li}_2\left[i\left(1-\sqrt{5}+\sqrt{5-2 \sqrt{5}}\right)\right]+\text{Li}_2\left[i\left(1+\sqrt{5}-\sqrt{5+2 \sqrt{5}}\right)\right] \right] = \dfrac{\pi}{5} \log 2 + K$$ using definition of Dilogarithm (pardon me if it's a trivial question, I haven't tried it yet). Jun 15, 2016 at 19:20
• Numerically your answer is off about 0.001 (according to W/A). Jun 17, 2016 at 4:52
• I evaluated numerically the integral. The result was $\mbox{int} = 1.35\color{#f00}{238704639191}$. Your result yields $\mbox{JDA} = K + \pi\ln\left(2\right)/5 \approx 1.35\color{#f00}{148281223794}$. Then, $$\mbox{int} - \mbox{JDA} = 0.00090423415397 \sim 10^{-4}$$ Jun 20, 2016 at 1:38

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

$$\quad$$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.

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 $$\pars{2}$$.

• What do you mean by 'direct numerical' calculation? Also, the answer is $$-\dfrac\pi5 \ln\left( 124 - 55\sqrt5 + 2\sqrt{7625 - 3410\sqrt5} \right) + \dfrac85 G$$ which matches numerically with your dilogarithm form. So can you tell how can we convert your form into the answer mathematically? Jul 1, 2016 at 11:02
• @IshanSingh Thanks for your remarks. $\color{#f00}{'Direct\ Numerical'}$: Just put the original integral in some CAS. I don't really know how to travel from 'my answer' to the one in your comment. Indeed, I would like to know it. Jul 1, 2016 at 15:40
• @IshanSingh . The dilogarithms can be rewritten as $4 \text{Ti}_2\left( \tan \left(\frac{ 3 \pi}{20} \right) \right)-4 \text{Ti}_2\left( \tan \left( \frac{ \pi}{20} \right) \right),$ where $\text{Ti}_2(x) = \Im \text{Li}_2(i x)$ is the inverse tangent integral. But since we have the identity $$\text{Ti}_2(\tan x) = x \ln \tan x +\sum_{n \geq 0} \frac{ \sin( 2 x (2n+1))}{(2n+1)^2}$$, and also $$\frac{ \tan \left( \frac{ \pi}{20} \right) }{ \tan^3 \left( \frac{ 3 \pi}{20} \right) } = 124 - 55 \sqrt{5} +2 \sqrt{7625 - 3410 \sqrt{5}},$$ it follow that (cont. next comment) Jul 3, 2016 at 11:02
• ... it follows that we only need to show that: $$\frac{2 G}{5} = \sum_{n=0}^{\infty} \frac{ \sin\left( \frac{3 \pi}{10} (2n+1)\right)-\sin\left( \frac{\pi}{10} (2n+1)\right)}{(2n+1)^2}.$$ I don't see how to prove it yet, but i'll let you know if i'll find out. Jul 3, 2016 at 11:05
• @nospoon We can prove the identity by showing that the numerator is periodic modulo 10. Sep 18, 2016 at 19:26