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I was motivated by the integral $\displaystyle \int_0^{\pi/12} \ln\tan(x)\,dx$ which was posted here which is equal to $-\frac{2}{3}G$ where is $G$ is Catalan's constant. With this motivation I came up with a result

$$\int_0^{\pi/24}\ln\tan x\,dx=\pi\ln\left(\frac{G\left(\frac{25}{24}\right)G\left(\frac{35}{24}\right)}{G\left(\frac{13}{24}\right)G\left(\frac{23}{24}\right)}\right)-\frac{\pi}{24}\ln\left(\frac{4096\pi^{12}}{\sqrt{2-\sqrt{3}}\left(2+\sqrt{2+\sqrt{3}}\right)^5}\right)$$Notation: $G$ denotes Barnes G-function.

Since Wolfram cannot generate the closed form however, the closed form obtained as per the WA check is found to be correct.

Moreover, I came up with following inequality, the close upper bound of the integral, surprisingly in terms of $ e$ and $G$ which is as follows.

$$\bigg|\int_0^{\pi/24}\ln\tan x \,dx\bigg| < \frac{e^{-1}}{G}\cdots (1)$$ Notation: $e$ denotes Euler's number.

Now I'm curious to know,

$\bullet$ Can we generalize the integral

$$\int_0^{\pi/p} \ln \tan x \,dx\, =\text{?} \; \; p\neq 0 $$

$\bullet$ How to prove $(1)$ using the inequality theorem?

Thank you

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  • $\begingroup$ See A class of series acceleration formulae for Catalan's constant, David M. Bradley, arxiv.org/pdf/0706.0356.pdf $\endgroup$ – FDP Jul 23 at 7:54
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Just approximations.

Concerning the integral

$$I(t)=\int_0^t \log (\tan (x))\,dx$$ $$I=\tan ^{-1}(\tan (t)) \log (\tan (t))+\frac{i}{2} (\text{Li}_2(i \tan (t))-\text{Li}_2(-i \tan (t)))$$

Since $t$ is supposed to be small, I should use $$I(t)=-t (1-\log (t))+\frac {t^2}9 \sum_{i=1}^\infty \frac{a_i}{b_i}\, t^{2i-1} $$ where the first coefficients are $$\left( \begin{array}{ccc} i & a_i & b_i \\ 1 & 1 & 1 \\ 2 & 7 & 50 \\ 3 & 62 & 2205 \\ 4 & 127 & 18900 \\ 5 & 146 & 81675 \\ 6 & 1414477 & 2766889125 \\ 7 & 32764 & 212837625 \\ 8 & 16931177 & 351486135000 \\ 9 & 11499383114 & 740606614122375 \\ 10 & 91546277357 & 17865510428390625 \end{array} \right)$$

Using $t=\frac \pi {24}$ this gives as a result $-0.39681136139214865267218614585540$ while the exact result should be $-0.39681136139214865267218614585537$

Edit

Another possibility is to let $\tan(x)=u$ to make $$I(t)=\int_0^{\tan ^{-1}(t)} \frac{\log (u)}{u^2+1}\,du= \tan ^{-1}(u)\log (u)+\frac{i}{2} (\text{Li}_2(i u)-\text{Li}_2(-i u))$$ and use $$\frac{i}{2} (\text{Li}_2(i u)-\text{Li}_2(-i u))=\sum_{n=0}^\infty \frac{(-1)^{n+1}}{(2 n+1)^2} u^n$$ which has the merit to be alternating.

More compact and exact will be $$\color{red}{I(t)=\tan ^{-1}\left(\tan ^{-1}(t)\right) \log \left(\tan ^{-1}(t)\right)-\frac{1}{4} \tan ^{-1}(t)\, \Phi \left(-\tan ^{-1}(t)^2,2,\frac{1}{2}\right)}$$ where appears the Lerch transcendent function.

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Only extended comment.

With CAS help:

$\int_0^{\frac{\pi }{p}} \ln (\tan (x)) \, dx=\tan ^{-1}\left(\tan \left(\frac{\pi }{p}\right)\right) \ln \left(\tan \left(\frac{\pi }{p}\right)\right)-\frac{1}{2} i \text{Li}_2\left(-i \tan \left(\frac{\pi }{p}\right)\right)+\frac{1}{2} i \text{Li}_2\left(i \tan \left(\frac{\pi }{p}\right)\right)$

for: $p > 2$, where: $\text{Li}_2(x)$ is polylogarithm function.

Mathematica code:

Integrate[Log[Tan[x]], {x, 0, Pi/p}] == ArcTan[Tan[\[Pi]/p]] Log[Tan[\[Pi]/p]] - 1/2 I PolyLog[2, -I Tan[\[Pi]/p]] + 1/2 I PolyLog[2, I Tan[\[Pi]/p]]]

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  • $\begingroup$ Or simpler: $-\frac{1}{2} i \text{Li}_2\left(-i \tan \left(\frac{\pi }{p}\right)\right)+\frac{1}{2} i \text{Li}_2\left(i \tan \left(\frac{\pi }{p}\right)\right)+\frac{\pi \ln \left(\tan \left(\frac{\pi }{p}\right)\right)}{p}$ $\endgroup$ – Mariusz Iwaniuk Jul 22 at 14:25
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Here is my attempt, however, it does not lead to a nice close form solution. Let $I_{p}$ be defined as follows:

\begin{equation} I_{p}=\int\limits_{0}^{\frac{\pi}{p}} \ln(\tan(x))\,dx \end{equation}

for some real valued $p$ such that $p\geq4$. With the substitution $x=\arctan(t)$, you can transform the integral to the following:

\begin{equation} I_{p}=\int\limits_{0}^{\tan(\pi/p)} \frac{\ln(t)}{1+t^{2}}\,dt \end{equation}

Now, by letting $t=e^{-z}$, you will get the following:

\begin{equation} I_{p}=\int\limits_{\ln(\cot(\pi/p))}^{+\infty} \frac{(-z)e^{-z}}{1+e^{-2z}}\,dz \end{equation}

\begin{equation} I_{p}=\int\limits_{\ln(\cot(\pi/p))}^{+\infty} \frac{(-z)e^{-z}}{1-(-e^{-2z})}\,dz \end{equation}

For simplicity, let $k=\ln(\cot(\pi/p))$. For any $p\geq4$, in the interval $[k,\infty)$, it holds that $0\leq|-e^{-2z}|\leq 1$, so it is justified to use the geometric series. Thus:

\begin{equation} I_{p}=-\int\limits_{k}^{+\infty} ze^{-z}\sum_{n=0}^{+\infty}(-e^{-2z})^{n}\,dz \end{equation}

\begin{equation} I_{p}=-\sum_{n=0}^{+\infty}(-1)^{n}\int\limits_{k}^{+\infty} ze^{-z}e^{-2nz}\,dz \end{equation}

\begin{equation} I_{p}=-\sum_{n=0}^{+\infty}(-1)^{n}\int\limits_{k}^{+\infty} ze^{-z(1+2n)}\,dz \end{equation}

By letting $s=z(1+2n)$, we obtain that:

\begin{equation} I_{p}=-\sum_{n=0}^{+\infty}\frac{(-1)^{n}}{(1+2n)^{2}}\int\limits_{k(1+2n)}^{+\infty} se^{-s}\,ds \end{equation}

Using the upper incomplete gamma function, we get the following:

\begin{equation} I_{p}=-\sum_{n=0}^{+\infty}\frac{(-1)^{n}}{(1+2n)^{2}}\Gamma(2,k+2nk) \end{equation}

\begin{equation} \boxed{\int\limits_{0}^{\frac{\pi}{p}} \ln(\tan(x))\,dx=-\sum_{n=0}^{+\infty}\frac{(-1)^{n}}{(1+2n)^{2}}\Gamma\left(2,\ln\left(\cot\left(\frac{\pi}{p}\right)\right)+2n\ln\left(\cot\left(\frac{\pi}{p}\right)\right)\right)} \end{equation}

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