TITLE: Integral $\int_0^{\large\frac{\pi}{4}}\left(\frac{1}{\log(\tan(x))}+\frac{1}{1-\tan(x)}\right)dx$
QUESTION [29 upvotes]: I am wondering if anyone would know how to evaluate this integral:
$$\int_{0}^{\Large\frac{\pi}{4}}\left(\frac{1}{\log(\tan(x))}+\frac{1}{1-\tan(x)}\right)dx.$$
I've tried, unsuccessfully, the change of variables $u=\tan (x)$.

REPLY [20 votes]: We have the following closed form evaluation.

$$
I:=\int_0^{\Large \frac{\pi}{4}}\!\!\left(\frac{1}{\ln(\tan x)}+\frac{1}{1-\tan x}\right)\! \mathrm dx= \color{blue}{\frac\pi8+\frac74\ln2+\frac\gamma2+\ln\pi-2\ln \Gamma\!\left(\!\frac14\!\right)} \tag1
$$

where $\gamma$ is the Euler-Mascheroni constant. 
A numerical approximation is
$$
I =\color{blue}{0.462999316582640135993449151416572....}
$$ 
As Lucian pointed out, by the change of variable $\displaystyle u=\tan x$, we readily have
$$
I=\int_0^1\left(\frac{1}{\ln u}+\frac{1}{1-u}\right)\frac{1}{1+u^2} \mathrm du.
$$
We set
$$
I(s):=\int_0^1\left(\frac{1}{\ln u}+\frac{1}{1-u}\right)\frac{u^s}{1+u^2} \mathrm du, \quad s>-1.
$$
The expected integral is thus $I(0)$.
Let's differentiate $I(s)$. 
We get 
$$
\begin{align}
I'(s)& =\int_0^1\left(\frac{1}{\ln u}+\frac{1}{1-u}\right)\frac{u^s\ln u}{1+u^2} \mathrm du \\\\
& =\int_0^1\left(1+\frac{\ln u}{1-u}\right)\frac{u^s}{1+u^2} \mathrm du \\\\
& =\int_0^1\! \frac{u^s}{1+u^2}\mathrm du +\int_0^1\!\frac{(1+u)u^s \ln u}{(1-u^2)(1+u^2)}\mathrm du \\\\
& =\int_0^1\! \frac{u^s(1-u^2)}{1-u^4}\mathrm du +\int_0^1\!\frac{u^s \ln u}{1-u^4}\mathrm du+\int_0^1\!\frac{u^{s+1} \ln u}{1-u^4}\mathrm du.
\end{align}
$$
By the change of variable $\displaystyle v=u^4$ in each of the preceding integrals and recalling the well known integral representations for the digamma function $\displaystyle \psi : = \Gamma'/\Gamma$ and for its derivative,
$$
\psi(s) = -\gamma+\int_0^1 \frac{1 - v^{s-1}}{1 -v}{\rm{d}} v, \quad s>0,
$$
$$
\psi'(s) = -\int_0^1 \frac{s^{s-1} \ln v}{1 - v}{\rm{d}} v, \quad s>0,
$$
we obtain

$$
I'(s)=\frac{1}{4}\psi\left(\frac{s+3}{4}\right)-\frac{1}{4}\psi\left(\frac{s+1}{4}\right)-\frac{1}{16}\psi'\left(\frac{s+2}{4}\right)-\frac{1}{16}\psi'\left(\frac{s+1}{4}\right). \tag2
$$ 

Since $$ \left|\left(\frac{1}{\ln u}+\frac{1}{1-u}\right)\frac{u^s}{1+u^2} \right| < u^s, \quad 0<u<1,\, s>-1,$$ 
giving
$$ |I(s)|\leq \int_0^1\left|\left(\frac{1}{\ln u}+\frac{1}{1-u}\right)\frac{u^s}{1+u^2} \right|\mathrm du < \!\!\int_0^1 u^s\mathrm du = \frac{1}{s+1}, $$ 
then, as $s \rightarrow +\infty$, we have $I(s) \rightarrow 0$.
We deduce that

$$
I(s)=\log \Gamma \left(\frac{s+3}{4}\right)\!-\!\log \Gamma \left(\frac{s+1}{4}\right) \!-\!\frac{1}{4}\psi\!\left(\frac{s+2}{4}\right)\!-\!\frac{1}{4}\psi\!\left(\frac{s+1}{4}\right)
, \, s>-1, \tag3
$$ 

and, for $s=0$,

$$
\begin{align}
I=\frac\pi8+\frac74\ln2+\frac\gamma2+\ln\pi-2\ln \Gamma\left(\frac14\right)
\end{align}
$$

where have used special values of the digamma function,
$$
\begin{align}
\psi \left(\frac12\right) & = -\gamma - 2\ln 2, \\
\psi \left(\frac14\right) & =  -\gamma + \frac\pi2- 3\ln 2,
\end{align}
$$
and the complement/reflection formula
$$
\Gamma\left(\frac34\right)\Gamma\left(\frac14\right)=\frac{\pi}{\sin(\frac\pi4)}=\pi \sqrt{2}.
$$