General integral $int_0^{frac{pi}{p}}lntan x ,dx $

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}=intlimits_{0}^{frac{pi}{p}} ln(tan(x)),dx end{equation}

for some real valued $p$ such that $pgeq4$. With the substitution $x=arctan(t)$, you can transform the integral to the following:

begin{equation} I_{p}=intlimits_{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}=intlimits_{ln(cot(pi/p))}^{+infty} frac{(-z)e^{-z}}{1+e^{-2z}},dz end{equation}

begin{equation} I_{p}=intlimits_{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 $pgeq4$, in the interval $[k,infty)$, it holds that $0leq|-e^{-2z}|leq 1$, so it is justified to use the geometric series. Thus:

begin{equation} I_{p}=-intlimits_{k}^{+infty} ze^{-z}sum_{n=0}^{+infty}(-e^{-2z})^{n},dz end{equation}

begin{equation} I_{p}=-sum_{n=0}^{+infty}(-1)^{n}intlimits_{k}^{+infty} ze^{-z}e^{-2nz},dz end{equation}

begin{equation} I_{p}=-sum_{n=0}^{+infty}(-1)^{n}intlimits_{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}}intlimits_{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{intlimits_{0}^{frac{pi}{p}} ln(tan(x)),dx=-sum_{n=0}^{+infty}frac{(-1)^{n}}{(1+2n)^{2}}Gammaleft(2,lnleft(cotleft(frac{pi}{p}right)right)+2nlnleft(cotleft(frac{pi}{p}right)right)right)} end{equation}

Only extended comment.

With CAS help:

$int_0^{frac{pi }{p}} ln (tan (x)) , dx=tan ^{-1}left(tan left(frac{pi }{p}right)right) ln left(tan left(frac{pi }{p}right)right)-frac{1}{2} i text{Li}_2left(-i tan left(frac{pi }{p}right)right)+frac{1}{2} i text{Li}_2left(i tan left(frac{pi }{p}right)right)$

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]]]

Just approximations.

Concerning the integral

$$I(t)=int_0^t log (tan (x)),dx$$ $$I=tan ^{-1}(tan (t)) log (tan (t))+frac{i}{2} (text{Li}_2(i tan (t))-text{Li}_2(-i tan (t)))$$

Since $t$ is supposed to be small, I should use $$I(t)=-t (1-log (t))+frac {t^2}9 sum_{i=1}^infty frac{a_i}{b_i}, t^{2i-1} $$ where the first coefficients are $$left( begin{array}{ccc} i & a_i & b_i \ 1 & 1 & 1 \ 2 & 7 & 50 \ 3 & 62 & 2205 \ 4 & 127 & 18900 \ 5 & 146 & 81675 \ 6 & 1414477 & 2766889125 \ 7 & 32764 & 212837625 \ 8 & 16931177 & 351486135000 \ 9 & 11499383114 & 740606614122375 \ 10 & 91546277357 & 17865510428390625 end{array} right)$$

Using $t=frac pi {24}$ this gives as a result $-0.39681136139214865267218614585540$ while the exact result should be $-0.39681136139214865267218614585537$

Edit

Another possibility is to let $tan(x)=u$ to make $$I(t)=int_0^{tan ^{-1}(t)} frac{log (u)}{u^2+1},du= tan ^{-1}(u)log (u)+frac{i}{2} (text{Li}_2(i u)-text{Li}_2(-i u))$$ and use $$frac{i}{2} (text{Li}_2(i u)-text{Li}_2(-i u))=sum_{n=0}^infty frac{(-1)^{n+1}}{(2 n+1)^2} u^n$$ which has the merit to be alternating.

More compact and exact will be $$color{red}{I(t)=tan ^{-1}left(tan ^{-1}(t)right) log left(tan ^{-1}(t)right)-frac{1}{4} tan ^{-1}(t), Phi left(-tan ^{-1}(t)^2,2,frac{1}{2}right)}$$ where appears the Lerch transcendent function.