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


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}

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,
$$
\color{#f00}{\int_{\pi/20}^{3\pi/20}\ln\pars{\tan\pars{x}}\,\dd x} =
\sum_{\omega_{k}}^{\infty}{\sin\pars{\omega_{k}\theta} \over \omega_{k}^{2}} -
\sum_{\omega_{k}}^{\infty}{\sin\pars{3\omega_{k}\theta} \over \omega_{k}^{2}}
$$

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.

\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 $\ds{G}$.

A: 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.
