Integral $\int_0^{\large\frac{\pi}{4}}\left(\frac{1}{\log(\tan(x))}+\frac{1}{1-\tan(x)}\right)dx$ – Calculus

Question

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 $u=\tan (x)$.

Answer 1

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 $$ \Gamma\left(\frac34\right)\Gamma\left(\frac14\right)=\frac{\pi}{\sin(\frac\pi4)}=\pi \sqrt{2}. $$