Prove $\int_{\frac{\pi}{20}}^{\frac{3\pi}{20}} \ln \tan x\,\,dx= - \frac{2G}{5}$

Question

Context:

This question asks to calculate a definite integral which turns out to be equal to $$\displaystyle 4 \, \text{Ti}_2\left( \tan \frac{3\pi}{20} \right) - 4 \, \text{Ti}_2\left( \tan \frac{\pi}{20} \right),$$ where $\text{Ti}_2(x) = \operatorname{Im}\text{Li}_2( i\, x)$ is the Inverse Tangent Integral function. The source for this integral is this question on brilliant.org. In a comment, the OP claims that the closed form can be further simplified to $-\dfrac\pi5 \ln\left( 124 - 55\sqrt5 + 2\sqrt{7625 - 3410\sqrt5} \right) + \dfrac85 G$.

How can we prove that?

I have thought about using the formula $$\text{Ti}_2(\tan x) = x \ln \tan x+ \sum_{n=0}^{\infty} \frac{\sin(2x(2n+1))}{(2n+1)^2}. \tag{1}$$ but that only mildly simplifies the problem.

Equivalent formulations include:

$$\, \text{Ti}_2\left( \tan \frac{3\pi}{20} \right) - \, \text{Ti}_2\left( \tan \frac{\pi}{20} \right) \stackrel?= \frac{ \pi}{20} \ln \frac{ \tan^3( 3\pi/20)}{\tan ( \pi/20)} + \frac{2 G}{5} \tag{2}$$

$$ \sum_{n=0}^{\infty} \frac{\sin \left(\frac{3\pi}{10}(2n+1) \right)- \sin \left(\frac{\pi}{10}(2n+1)\right)}{(2n+1)^2} \stackrel?=\ \frac{2G}{5} \tag{3}$$ $$\int_{\pi/20}^{3\pi/20} \ln \tan x\,\,dx \stackrel?= - \frac{2G}{5} \tag{4}$$

A related similar question is this one.

Answer 1

We can use the same technique of the linked answer for proving the claim. We first prove that $$2\int_{0}^{\pi/20}\log\left(\tan\left(5x\right)\right)dx=\int_{\pi/20}^{3\pi/20}\log\left(\tan\left(x\right)\right)dx.\tag{1}$$Let $$I=\int_{0}^{\pi/20}\log\left(\tan\left(5x\right)\right)dx$$ using the identity

$$\tan\left(\left(2n+1\right)x\right)=\tan(x)\prod_{k=1}^{n}\tan\left(\frac{k\pi}{2n+1}+x\right)\tan\left(\frac{k\pi}{2n+1}-x\right)$$

we have $$\begin{align}I= & \int_{0}^{\pi/20}\log\left(\tan\left(x\right)\right)dx+\int_{0}^{\pi/20}\log\left(\tan\left(\frac{\pi}{5}-x\right)\right)dx \\ + & \int_{0}^{\pi/20}\log\left(\tan\left(\frac{\pi}{5}+x\right)\right)dx+\int_{0}^{\pi/20}\log\left(\tan\left(\frac{2\pi}{5}-x\right)\right)dx \\ + & \int_{0}^{\pi/20}\log\left(\tan\left(\frac{2\pi}{5}+x\right)\right)dx \\ = & \int_{0}^{\pi/20}\log\left(\tan\left(x\right)\right)dx+\int_{3\pi/20}^{\pi/5}\log\left(\tan\left(x\right)\right)dx \\ + &\int_{\pi/5}^{\pi/4}\log\left(\tan\left(x\right)\right)dx+\int_{7\pi/20}^{2\pi/5}\log\left(\tan\left(x\right)\right)dx \\ + &\int_{2\pi/5}^{9\pi/20}\log\left(\tan\left(x\right)\right)dx. \end{align} $$ So we have $$I=\int_{0}^{\pi/20}\log\left(\tan\left(x\right)\right)dx+\int_{3\pi/20}^{\pi/4}\log\left(\tan\left(x\right)\right)dx+\int_{7\pi/20}^{9\pi/20}\log\left(\tan\left(x\right)\right)dx\tag{2} $$ and in the last integral of $(2)$ if we put $x\rightarrow\frac{\pi}{2}-x $ and recalling the identity $\tan\left(\frac{\pi}{2}-x\right)=\frac{1}{\tan\left(x\right)} $, we get $$\begin{align}I= & \int_{0}^{\pi/20}\log\left(\tan\left(x\right)\right)dx+\int_{3\pi/20}^{\pi/4}\log\left(\tan\left(x\right)\right)dx-\int_{\pi/20}^{3\pi/20}\log\left(\tan\left(x\right)\right)dx \\ = & \int_{0}^{\pi/4}\log\left(\tan\left(x\right)\right)-2\int_{\pi/20}^{3\pi/20}\log\left(\tan\left(x\right)\right)dx \\ = & 5\int_{0}^{\pi/20}\log\left(\tan\left(5x\right)\right)dx-2\int_{\pi/20}^{3\pi/20}\log\left(\tan\left(x\right)\right)dx \end{align}$$ so finally we have $(1)$. Hence $$\begin{align} \int_{\pi/20}^{3\pi/20}\log\left(\tan\left(x\right)\right)dx= & 2\int_{0}^{\pi/20}\log\left(\tan\left(5x\right)\right)dx \\ = & \frac{2}{5}\int_{0}^{\pi/4}\log\left(\tan\left(x\right)\right)dx \\ = & \color{red}{-\frac{2}{5}G} \end{align}$$ as wanted.

Answer 2

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

There is a well know series for $\ds{\ln\pars{\tan\pars{x}}}$. Namely \begin{align} \ln\pars{\tan\pars{x}} & = -2\sum_{k = 0}^{\infty}{\cos\pars{2\bracks{2k + 1}x} \over 2k + 1}\,,\qquad x\ \in\ \pars{0,{\pi \over 2}} \end{align}

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Then, \begin{align} \color{#f00}{\int_{\pi/20}^{3\pi/20}\ln\pars{\tan\pars{x}}\,\dd x} & = -2\sum_{k = 0}^{\infty}{1 \over 2k + 1} \int_{\pi/20}^{3\pi/20}\cos\pars{2\bracks{2k + 1}x}\,\dd x \\[3mm] &= -2\sum_{k = 0}^{\infty}{1 \over 2k + 1} \int_{-\pi/20}^{\pi/20}\cos\pars{2\bracks{2k + 1}\pars{x + \pi/10}}\,\dd x \\[3mm] &= -4\sum_{k = 0}^{\infty}{\cos\pars{\bracks{2k + 1}\pi/5} \over 2k + 1} \int_{0}^{\pi/20}\cos\pars{2\bracks{2k + 1}x}\,\dd x \\[3mm] &= -2\sum_{k = 0}^{\infty}{% \sin\pars{\bracks{2k + 1}\pi/10}\cos\pars{\bracks{2k + 1}\pi/5} \over \pars{2k + 1}^{2}} \end{align}

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Preliminary result: $$ \color{#f00}{\int_{\pi/20}^{3\pi/20}\ln\pars{\tan\pars{x}}\,\dd x} = -2\sum_{\omega_{k}}^{\infty}{% \sin\pars{\omega_{k}\theta}\cos\pars{2\omega_{k}\theta} \over \omega_{k}^{2}}\,,\qquad \left\lbrace\begin{array}{rcl} \ds{\omega_{k}} & \ds{\equiv} & \ds{2k + 1\,,\quad k = 0,1,2,\ldots} \\[1mm] \ds{\theta} & \ds{\equiv} & \ds{\pi \over 10} \end{array}\right. $$ Moreover, $$ \color{#f00}{\int_{\pi/20}^{3\pi/20}\ln\pars{\tan\pars{x}}\,\dd x} = \sum_{\omega_{k}}^{\infty}{\sin\pars{\omega_{k}\theta} \over \omega_{k}^{2}} - \sum_{\omega_{k}}^{\infty}{\sin\pars{3\omega_{k}\theta} \over \omega_{k}^{2}} $$

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Note that \begin{align} \sum_{\omega_{k}}{\sin\pars{\omega_{k}t} \over \omega_{k}^{2}} & = \Im\sum_{k = 1}^{\infty}{\expo{\ic\pars{2k + 1}t} \over \pars{2k + 1}^{2}} = \half\,\Im\sum_{k = 1}^{\infty}{\pars{\expo{\ic t}}^{k} \over k^{2}} - \half\,\Im\sum_{k = 1}^{\infty}{\pars{-\expo{\ic t}}^{k} \over k^{2}} \\[3mm] & = \half\,\Im\Li{2}\pars{\expo{\ic t}} - \half\,\Im\Li{2}\pars{-\expo{\ic t}} \end{align} Note that $\ds{\Li{2}\pars{z} - \Li{2}\pars{-z} = 2\,\chi_{2}\pars{z} = 2\sum_{k = 0}^{\infty}{z^{2k + 1} \over \pars{2k + 1}^{2}}}$ where $\ds{\chi}$ is the Legendre Chi Function.

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\begin{align} \color{#f00}{\int_{\pi/20}^{3\pi/20}\ln\pars{\tan\pars{x}}\,\dd x} & = \color{#f00}{\half\,\Im\Li{2}\pars{\expo{\ic\pi/10}} - \half\,\Im\Li{2}\pars{-\expo{\ic\pi/10}}} \\ & - \color{#f00}{\half\,\Im\Li{2}\pars{\expo{3\ic\pi/10}} + \half\,\Im\Li{2}\pars{-\expo{3\ic\pi/10}}} \end{align}

I'm still trying to find the relation with the Catalan Constant $\ds{G}$.

Answer 3

Let $J(a)=\int_{\frac{\pi}{20}}^{\frac{3\pi}{20}}\tanh^{-1}\frac{2a\cos2x}{1+a^{2}}dx$ and evaluate
\begin{align} J’(a) &= \int_{\frac\pi{20}}^{\frac{3\pi}{20}}\frac{2(1-a^2)\cos2x}{a^4+1-2a^2\cos4x}dx=\frac1{2a}{\left.\tan^{-1}\frac{2a\sin2x}{1-a^2}\right|_{\frac{\pi}{20}}^{\frac{3\pi}{20} } }\\ &=\frac1{2a}\tan^{-1}\frac{a-a^5}{1+a^6} = \frac1{2a}(\tan^{-1}a - \tan^{-1}a^5) \end{align} where $\sin\frac{3\pi}{10}-\sin\frac{\pi}{10}=\frac12$ and $\sin\frac{3\pi}{10}\sin\frac{\pi}{10}=\frac14$ are recognized. Then \begin{align} \int_\frac\pi{20}^\frac{3\pi}{20}\ln(\tan x)~dx &= -\int_\frac\pi{20}^\frac{3\pi}{20}\tanh^{-1}(\cos2x)dx =-J(1) =-\int_0^1 J’(a)da \\ & =-\frac12 \int_0^1\left(\frac{\tan^{-1}a}{a}\right. -\underset{a^5\to a}{\left.\frac{\tan^{-1}a^5}{a}\right)}da\\ &=-\left(\frac12-\frac1{10}\right) \int_0^1\frac{\tan^{-1}a}{a}da=-\frac25 G \end{align}