I just wonder if this integral is correct.

How can we show that?

$$\int_{0}^{\pi/2}\left(\frac{1}{\ln\tan x}+\frac{2-\sqrt[3]{\tan^4x}}{1-\tan x}\right)\mathrm dx=\color{blue}\pi$$

Let $u=\tan x$

$$\int_{0}^{\infty}\left(\frac{1}{\ln u}+\frac{2-u^{4/3}}{1-u}\right)\frac{\mathrm du}{1+u^2}$$

Let $u=e^{y}$

$$\int_{1}^{\infty}\left(\frac{1}{y}+\frac{2-e^{4y/3}}{1-e^{y}}\right)\frac{e^y\mathrm du}{1+e^{2y}}$$

$$\frac{1}{2}\int_{-\infty}^{\infty}\left(\frac{1}{y}+\frac{2-e^{4y/3}}{1-e^{y}}\right)\frac{\mathrm dy}{\cosh y}$$

I can't continue...

  • 1
    $\begingroup$ The final integral (in $dy$, not $du$) should run from $-\infty$ to $\infty$, since $e^y=u=0$ implies $y=-\infty$, not $y=1$. It should also have a $1\over2$, since $\cosh y=(e^y+e^{-2})/2$. $\endgroup$ – Barry Cipra May 7 '18 at 19:01
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    $\begingroup$ from where do you know this result? $\endgroup$ – Dr. Sonnhard Graubner May 7 '18 at 19:02
  • $\begingroup$ hey bui, how are you. still not givin up? :-) $\endgroup$ – tired May 7 '18 at 22:02
  • $\begingroup$ hi @tired how do you know it is me? I am very proud of you!! Where is that devil. What it is his name again? Please remind me!! $\endgroup$ – user516887 May 7 '18 at 22:46
  • $\begingroup$ in germany there is a saying, which roughly translates as: "the devil is a squirrel"... $\endgroup$ – tired May 7 '18 at 23:29

Let $I$ be given by the integral


Enforce the substitution $x\mapsto \arctan(x)$ in $(1)$ to reveal

$$\begin{align} I&=\int_0^\infty \left(\frac{1}{\log(x)}-\frac{2-x^{4/3}}{x-1}\right)\,\frac{1}{1+x^2}\,dx\\\\ &=\int_0^\infty \left(\frac{1}{\log(x)}-\frac{1}{x-1}\right)\,\frac{1}{1+x^2}\,dx+\int_0^\infty \left(\frac{1-x^{4/3}}{1-x}\right)\,\frac{1}{1+x^2}\,dx\tag2 \end{align}$$

The first integral on the right-hand side of $(2)$ is easy to evaluate by symmetry. Let $J$ be given by

$$J=\int_0^\infty \left(\frac{1}{\log(x)}-\frac{1}{x-1}\right)\,\frac{1}{1+x^2}\,dx\tag3$$

and enforce the substitution $x\mapsto 1/x$. Then, we find that

$$J=\int_0^\infty \left(-\frac{1}{\log(x)}-\frac{x}{1-x}\right)\,\frac{1}{1+x^2}\,dx\tag4$$

Adding $(3)$ and $(4)$, we find that

$$2J=\int_0^\infty \frac{1}{1+x^2}\,dx=\frac\pi2$$

Hence, $J=\frac\pi4$.

We now proceed to evaluate the second integral on the right-hand side of $(2)$. To that end, let $K$ be the integral given by

$$\begin{align} K&=\int_0^\infty \left(\frac{1-x^{4/3}}{1-x}\right)\,\frac{1}{1+x^2}\,dx\\\\ &\overbrace{=}^{x\mapsto x^3}3\int_0^\infty \frac{1-x^4}{1-x^3}\frac{x^2}{1+x^6}\,dx\\\\ &=3\int_0^\infty \frac{x^2(1+x+x^2+x^3)}{(1+x+x^2)(1+x^6)}\,dx\\\\ &\overbrace{=}^{x\mapsto 1/x}3\int_0^\infty \frac{x(1+x+x^2+x^3)}{(1+x+x^2)(1+x^6)}\,dx\\\\ &=3\int_0^\infty \frac{x(x+1)}{(x^2+x+1)(x^4-x^2+1)}\tag5\\\\ &=\frac{3\pi}{4} \end{align}$$

where $(5)$ can be evaluated using partial fraction expansion.

Putting it together yields the coveted equality


  • $\begingroup$ Thank you @Mark Viola. $\endgroup$ – user516887 May 7 '18 at 21:48
  • $\begingroup$ You're welcome! My pleasure. $\endgroup$ – Mark Viola May 7 '18 at 21:55

Elementary manipulations with the classical beta function reveal that

$$I=\frac{\pi}{2}\int_0^1 \frac{2+\cos(\pi x/2)+3\sin(\pi x/2)}{1+\cos(\pi x/2)+\sin(\pi x/2)}\textrm{d} x =\pi.$$

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    $\begingroup$ where on earth did you learn this dark wizardry from! $\endgroup$ – frogeyedpeas May 7 '18 at 19:59
  • $\begingroup$ How did you arrive at this expression? $\endgroup$ – Mark Viola May 7 '18 at 20:22
  • 3
    $\begingroup$ To what "elementary manipulations" do you refer? This is not a useful solution without showing some intermediate steps. Otherwise, why not simply write "Elementary manipulations with the classical beta function reveal that $I=\pi$?" Would that be useful? $\endgroup$ – Mark Viola May 7 '18 at 21:03

Under $x\to\frac{\pi}{2}-x$, one has \begin{eqnarray} I&=&\int_{0}^{\pi/2}\left(\frac{1}{\ln\tan x}+\frac{2-\sqrt[3]{\tan^4x}}{1-\tan x}\right)\mathrm dx\tag{1}\\ &=&\int_{0}^{\pi/2}\left(\frac{1}{\ln\cot x}+\frac{2-\sqrt[3]{\cot^4x}}{1-\cot x}\right)\mathrm dx\\ &=&\int_{0}^{\pi/2}\left(-\frac{1}{\ln\tan x}+\frac{\frac1{\sqrt[3]{\tan x}}-2\tan x}{1-\tan x}\right)\mathrm dx.\tag{2} \end{eqnarray} Adding (1) to (2), one has \begin{eqnarray} 2I&=&\int_{0}^{\pi/2}\frac{\bigg(\frac1{\sqrt[3]{\tan x}}-\sqrt[3]{\tan^4x}\bigg)+2(1-\tan x)}{1-\tan x}\mathrm dx\\ &=&\pi+\int_{0}^{\pi/2}\frac{\frac1{\sqrt[3]{\tan x}}-\sqrt[3]{\tan^4x}}{1-\tan x}\mathrm dx \end{eqnarray} and the rest is the same as Mark Viola's answer under $u=\sqrt[3]{\tan x}$.


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