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

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

Answer 2

Noti ce that this is not an answer with same limits as question, but I'm leaving it here for others to help them if the upper limit was changed.

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We have the following integral: $$I=\int_{0}^{\Large\frac{\pi}{2}}\left(\frac{1}{\log(\tan(x))}+\frac{1}{1-\tan(x)}\right)dx\tag{$I$}$$ Use $\displaystyle \int_0^a f(x) dx=\int_0^a f(a-x) dx$ $$I=\int_{0}^{\Large\frac{\pi}{2}}\left(\frac{1}{\log(\tan(\pi/2-x))}+\frac{1}{1-\tan(\pi/2-x)}\right)dx$$ Use $\displaystyle \tan(\pi/2-x)=\cot(x)$ $$ I=\int_{0}^{\Large\frac{\pi}{2}}\left(\frac{1}{\log(\cot(x))}+\frac{1}{1-\cot(x)}\right)dx\tag{$II$}$$ Add $(I)$ and $(II)$ $$2I=\int_{0}^{\Large\frac{\pi}{2}}\left(\frac{1}{\log(\tan(x))}+\frac{1}{1-\tan(x)}\right)dx+\int_{0}^{\Large\frac{\pi}{2}}\left(\frac{1}{\log(\cot(x))}+\frac{1}{1-\cot(x)}\right)dx$$ Simplify using $\displaystyle \log(\cot x)=\log(1/\tan x)=-\log(\tan x)$ $$2I=\int_{0}^{\Large\frac{\pi}{2}}\left(\frac{1}{1-\cot(x)}+\frac{1}{1-\tan x}\right)dx$$ Use $\displaystyle \cot x\tan x=1$ after taking common denominators to get: $$2I=\int_{0}^{\Large\frac{\pi}{2}}dx=\frac{\pi}2$$ Just the final step: $$I=\frac{\pi}4\Box$$

Answer 3

This is my friend’s solution. He doesn’t use Math Stack Exchange, so I’m posting it on his behalf.

Let $$ I = \int_0^{\pi/4} \left( \frac{1}{\ln(\tan x)} + \frac{1}{1 - \tan x} \right) dx. $$

Set $t = \tan x$, so $x \in [0, \pi/4]$ corresponds to $t \in [0, 1]$ and $dx = \frac{dt}{1 + t^2}$. Then:

$$ I = \int_0^1 \left( \frac{1}{(1 + t^2)\ln t} + \frac{1}{(1 + t^2)(1 - t)} \right) dt = \int_0^1 \frac{1}{1 + t^2} \left( \frac{1}{\ln t} + \frac{1}{1 - t} \right) dt. $$

Break this into two integrals:

$$ I_1 = \int_0^1 \frac{dt}{(1 + t^2)\ln t}, \qquad I_2 = \int_0^1 \frac{dt}{(1 + t^2)(1 - t)}. $$

Use the substitution $t = e^{-s}$ in each integral. Then $dt = -e^{-s} ds$, $t = 1$ corresponds to $s = 0$, and $t \to 0^+$ corresponds to $s \to \infty$. One finds:

$$ I_1 = \int_{\infty}^0 \frac{-e^{-s}}{(1 + e^{-2s})(-s)},ds = \int_0^\infty \frac{e^{-s}}{s(1 + e^{-2s})} ds = \int_0^\infty \frac{ds}{2s\cosh s}, $$

since $1 + e^{-2s} = 2e^{-s} \cosh s$.

Similarly,

$$ I_2 = \int_{\infty}^0 \frac{-e^{-s}}{(1 + e^{-2s})(1 - e^{-s})} ds = \int_0^\infty \frac{e^{-s}}{(1 - e^{-s})(1 + e^{-2s})} ds = \int_0^\infty \frac{ds}{2\cosh s (1 - e^{-s})}. $$

Hence,

$$ I = I_1 + I_2 = \frac{1}{2} \int_0^\infty \frac{1}{\cosh s} \left( \frac{1}{s} + \frac{1}{1 - e^{-s}} \right) ds. $$

Combine the bracketed terms using the expansion:

$$ \frac{1}{1 - e^{-s}} - \frac{1}{s} = \frac{1}{2} + \frac{s}{12} - \frac{s^3}{720} + \cdots, $$

then integrate termwise against $\frac{1}{2\cosh s}$. In particular, use the known series expansions:

$$ \int_0^\infty \frac{s^n}{\cosh s} ds = 2,n! \sum_{k=0}^\infty \frac{(-1)^k}{(2k + 1)^{n+1}} = 2,n! ,\beta(n+1), $$

where $\beta$ is the Dirichlet beta function. After simplifying, one eventually arrives at:

$$ I = \frac{1}{2} - \beta’(0) - \frac{1}{4} \left[ \psi\left( \frac{1}{4} \right) + \psi\left( \frac{1}{2} \right) \right]. $$

Using the known values:

$$ \beta’(0) = \ln\left( \frac{\Gamma(1/4)^2}{2\pi \sqrt{2}} \right), \quad \psi\left( \frac{1}{2} \right) = -\gamma - 2\ln 2, \quad \psi\left( \frac{1}{4} \right) = -\gamma - \frac{\pi}{2} - 3\ln 2, $$

we simplify:

$$ I = \frac{\pi}{8} + \frac{\gamma}{2} + \ln\left( \frac{2^{7/4} \pi}{\Gamma(1/4)^2} \right). $$