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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 $u=\tan (x)$.

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    $\begingroup$ For the second half use, $$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$ $\endgroup$ – lab bhattacharjee Aug 29 '14 at 16:31
  • $\begingroup$ That's what I did too, @labbhattacharjee. What would you recommend for the first term in the integrand? I tried the same mapping, $x \mapsto \frac{\pi}{4} - x$, but I ended up with $$\int_{0}^{\frac{\pi}{4}} \dfrac{1}{\log(1-\tan x) - \log(1+\tan x)}$$. $\endgroup$ – Kari Aug 29 '14 at 16:41
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    $\begingroup$ Khallil, you can write the difference of logs as a log of a quotient, and then apply the tangent angle sum identity. $\endgroup$ – Travis Aug 29 '14 at 16:44
  • $\begingroup$ Do you know the source for this integral? $\endgroup$ – Ali Caglayan Aug 29 '14 at 16:50
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    $\begingroup$ @Aditya: I am afraid that the OP is genuinely interested in the $\dfrac\pi4$ case, inasmuch as, if we were to substitute $u=\tan x$, the integral would become $\displaystyle\int_0^1\bigg(\dfrac1{\ln x}+\dfrac1{1-x}\bigg)\dfrac{dx}{1+x^2}$, which looks awfully similar to $\displaystyle\int_0^1\bigg(\dfrac1{\ln x}+\dfrac1{1-x}\bigg)~dx=\gamma$, where $\gamma$ is the Euler-Mascheroni constant. $\endgroup$ – Lucian Aug 29 '14 at 18:16
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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 $$ \Gamma\left(\frac34\right)\Gamma\left(\frac14\right)=\frac{\pi}{\sin(\frac\pi4)}=\pi \sqrt{2}. $$

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  • $\begingroup$ Can one safely say, then, that this cannot integrate to an elementary function? $\endgroup$ – Addem Aug 30 '14 at 6:41
  • $\begingroup$ @OlivierOloa Thank you! $\endgroup$ – Euler's student Sep 1 '14 at 15:59
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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.


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: $$I=\frac{\pi}4\Box$$

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  • $\begingroup$ You seem to have confused $\frac{\pi}{4}$ and $\frac{\pi}{2}$ in places. $\endgroup$ – Travis Aug 29 '14 at 17:06
  • $\begingroup$ @Travis edited /. $\endgroup$ – RE60K Aug 29 '14 at 17:07
  • $\begingroup$ The upper bound of the integral is $\frac{\pi}{4}$, @Aditya. $\endgroup$ – Kari Aug 29 '14 at 17:10
  • $\begingroup$ The limits on the original integral were $0$ and $\frac{\pi}{4}$. (NB that the (improper) integral of the first term over $[0, \frac{\pi}{2}]$ is just zero.) $\endgroup$ – Travis Aug 29 '14 at 17:10
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    $\begingroup$ I believe that it probably is $\dfrac\pi4$, not $\dfrac\pi2$. In which case, this answer is the correct one. Olivier gets $\displaystyle\frac\pi8+\frac74\ln2+\frac\gamma2+\ln\pi-2\ln \Gamma\left(\frac14\right)$ as the answer. $\endgroup$ – Akiva Weinberger Aug 29 '14 at 23:23

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