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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.

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2 Answers 2

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

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  • $\begingroup$ 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? $\endgroup$
    – MathGod
    Jun 15, 2016 at 17:08
  • $\begingroup$ @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 $\frac{\theta}{\sin\theta}$. $\endgroup$ Jun 15, 2016 at 17:20
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    $\begingroup$ 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). $\endgroup$
    – MathGod
    Jun 15, 2016 at 19:20
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    $\begingroup$ Numerically your answer is off about 0.001 (according to W/A). $\endgroup$
    – MathGod
    Jun 17, 2016 at 4:52
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    $\begingroup$ 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, $$ \mbox{int} - \mbox{JDA} = 0.00090423415397 \sim 10^{-4} $$ $\endgroup$ Jun 20, 2016 at 1:38
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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $$ \int_{0}^{1} \ln\pars{x^{2} - 2x - 4 \over x^{2} + 2x -4}\, {\dd x \over \root{1 -x^{2}}} =\, {\Large ?} $$

$\quad$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 $\pars{2}$.

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  • $\begingroup$ 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? $\endgroup$
    – MathGod
    Jul 1, 2016 at 11:02
  • $\begingroup$ @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. $\endgroup$ Jul 1, 2016 at 15:40
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    $\begingroup$ @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) $\endgroup$ Jul 3, 2016 at 11:02
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    $\begingroup$ ... 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. $\endgroup$ Jul 3, 2016 at 11:05
  • $\begingroup$ @nospoon We can prove the identity by showing that the numerator is periodic modulo 10. $\endgroup$
    – MathGod
    Sep 18, 2016 at 19:26

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