General integral $\int_0^{\frac{\pi}{p}}\ln\tan x \,dx $ I was motivated by the integral  $\displaystyle \int_0^{\pi/12} \ln\tan(x)\,dx$ which was posted here which is equal to $-\frac{2}{3}G$ where is $G$ is Catalan's constant.
With this motivation I came up with a result

$$\int_0^{\pi/24}\ln\tan x\,dx=\pi\ln\left(\frac{G\left(\frac{25}{24}\right)G\left(\frac{35}{24}\right)}{G\left(\frac{13}{24}\right)G\left(\frac{23}{24}\right)}\right)-\frac{\pi}{24}\ln\left(\frac{4096\pi^{12}}{\sqrt{2-\sqrt{3}}\left(2+\sqrt{2+\sqrt{3}}\right)^5}\right)$$Notation: $G$ denotes Barnes G-function.

Since Wolfram cannot generate the closed form however, the closed form obtained as  per the WA check is found to be correct.
Moreover, I came up with following inequality, the close upper bound of the integral, surprisingly in terms of $ e$ and $G$ which is as follows.

$$\bigg|\int_0^{\pi/24}\ln\tan x \,dx\bigg|  < \frac{e^{-1}}{G}\cdots (1)$$
Notation: $e$ denotes  Euler's number.

Now I'm curious to know,
$\bullet$ Can we generalize the integral

$$\int_0^{\pi/p} \ln \tan x \,dx\, =\text{?} \; \; p\neq  0 $$

$\bullet$ How to prove  $(1)$ using the inequality theorem?
Thank you
 A: Only extended comment.
With CAS help:
$\int_0^{\frac{\pi }{p}} \ln (\tan (x)) \, dx=\tan ^{-1}\left(\tan \left(\frac{\pi }{p}\right)\right) \ln \left(\tan \left(\frac{\pi }{p}\right)\right)-\frac{1}{2} i \text{Li}_2\left(-i \tan \left(\frac{\pi
   }{p}\right)\right)+\frac{1}{2} i \text{Li}_2\left(i \tan \left(\frac{\pi }{p}\right)\right)$
for: $p > 2$, where: $\text{Li}_2(x)$ is  polylogarithm function.
Mathematica code:
Integrate[Log[Tan[x]], {x, 0, Pi/p}] == ArcTan[Tan[\[Pi]/p]] Log[Tan[\[Pi]/p]] -  1/2 I PolyLog[2, -I Tan[\[Pi]/p]] + 1/2 I PolyLog[2, I Tan[\[Pi]/p]]]
A: Here is my attempt, however, it does not lead to a nice close form solution. Let $I_{p}$ be defined as follows:
\begin{equation}
I_{p}=\int\limits_{0}^{\frac{\pi}{p}} \ln(\tan(x))\,dx
\end{equation}
for some real valued $p$ such that $p\geq4$. With the substitution $x=\arctan(t)$, you can transform the integral to the following:
\begin{equation}
I_{p}=\int\limits_{0}^{\tan(\pi/p)} \frac{\ln(t)}{1+t^{2}}\,dt
\end{equation}
Now, by letting $t=e^{-z}$, you will get the following:
\begin{equation}
I_{p}=\int\limits_{\ln(\cot(\pi/p))}^{+\infty} \frac{(-z)e^{-z}}{1+e^{-2z}}\,dz
\end{equation}
\begin{equation}
I_{p}=\int\limits_{\ln(\cot(\pi/p))}^{+\infty} \frac{(-z)e^{-z}}{1-(-e^{-2z})}\,dz
\end{equation}
For simplicity, let $k=\ln(\cot(\pi/p))$. For any $p\geq4$, in the interval $[k,\infty)$, it holds that $0\leq|-e^{-2z}|\leq 1$, so it is justified to use the geometric series. Thus:
\begin{equation}
I_{p}=-\int\limits_{k}^{+\infty} ze^{-z}\sum_{n=0}^{+\infty}(-e^{-2z})^{n}\,dz
\end{equation}
\begin{equation}
I_{p}=-\sum_{n=0}^{+\infty}(-1)^{n}\int\limits_{k}^{+\infty} ze^{-z}e^{-2nz}\,dz
\end{equation}
\begin{equation}
I_{p}=-\sum_{n=0}^{+\infty}(-1)^{n}\int\limits_{k}^{+\infty} ze^{-z(1+2n)}\,dz
\end{equation}
By letting $s=z(1+2n)$, we obtain that:
\begin{equation}
I_{p}=-\sum_{n=0}^{+\infty}\frac{(-1)^{n}}{(1+2n)^{2}}\int\limits_{k(1+2n)}^{+\infty} se^{-s}\,ds
\end{equation}
Using the upper incomplete gamma function, we get the following:
\begin{equation}
I_{p}=-\sum_{n=0}^{+\infty}\frac{(-1)^{n}}{(1+2n)^{2}}\Gamma(2,k+2nk)
\end{equation}
\begin{equation}
\boxed{\int\limits_{0}^{\frac{\pi}{p}} \ln(\tan(x))\,dx=-\sum_{n=0}^{+\infty}\frac{(-1)^{n}}{(1+2n)^{2}}\Gamma\left(2,\ln\left(\cot\left(\frac{\pi}{p}\right)\right)+2n\ln\left(\cot\left(\frac{\pi}{p}\right)\right)\right)}
\end{equation}
