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I am trying to evaluate: $$\int_0^{\pi/12} \ln(\tan x)\,dx$$

I think the integral is quite simple but I am having a hard time evaluating it. I started with the result: $$\int_0^{\pi/4} \ln(\tan x)\,dx= -G$$ where $G$ is the Catalan's constant. With the change of variables $x\rightarrow 3x$ and using the fact that $\tan(3x)=\tan x\tan\left(\frac{\pi}{3}+x\right)\tan\left(\frac{\pi}{3}-x\right)$, the integral is: $$\int_0^{\pi/12}\ln(\tan x)\,dx+\int_0^{\pi/12}\ln \tan\left(\frac{\pi}{3}+x\right)\,dx+\int_0^{\pi/12}\ln \tan\left(\frac{\pi}{3}-x\right)\,dx=-\frac{G}{3}$$ $$\Rightarrow \int_0^{\pi/12}\ln(\tan x)\,dx+\int_{-\pi/12}^{\pi/12}\ln \tan\left(\frac{\pi}{3}+x\right)\,dx=-\frac{G}{3}$$ But I do not see how to proceed.

Help is appreciated. Thanks!

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    $\begingroup$ perhaps you know this, $\log \sin x$ is Lobachevsky's function, no elementary indefinite integral. As always, this may not prevent a definite integral on lucky endpoints from having a value that can be calculated... $\endgroup$ – Will Jagy Oct 20 '14 at 19:30
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    $\begingroup$ $$\int_0^{\pi/12} \ln(\tan x)\,dx=-\frac{4G}{3}-\frac{1}{12}\bigg[5\pi\ln\big(2+\sqrt{3}\big)-6i\,\text{Li}_2\big(-i(2+ \sqrt{3})\big)-\text{Li}_2\big(i(2+ \sqrt{3})\big)\bigg]$$ $\endgroup$ – Venus Oct 20 '14 at 19:41
  • $\begingroup$ @Venus: Is it possible to simplify the dilogs and how do you arrive at that expression? Thanks! $\endgroup$ – Pranav Arora Oct 20 '14 at 19:44
  • $\begingroup$ @PranavArora The real and imaginary parts of those dilogarithmic terms are the topic of this question. So possibly. $\endgroup$ – David H Oct 20 '14 at 19:46
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    $\begingroup$ $$\int_0^z\ln(\tan x)\,dx=z\,\ln(\tan z)-\operatorname{Ti}_2(\tan z)$$ $\endgroup$ – Vladimir Reshetnikov Jan 3 '16 at 20:50
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First: $~\displaystyle 2\int_0^{\tfrac{\pi}{12}} \log(\tan(3x))dx=\int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx\qquad(1)$

Proof:

Let $I=\displaystyle \int_0^{\tfrac{\pi}{12}} \log(\tan(3x))dx$

$\tan(3x)=\tan(x)\tan\big(\dfrac{\pi}{3}+x\big)\tan\big(\dfrac{\pi}{3}-x\big)$

$\displaystyle I= \int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx+\int_0^{\tfrac{\pi}{12}} \log\Big(\tan\Big (\dfrac{\pi}{3}+x\Big)\Big)dx+\int_0^{\tfrac{\pi}{12}} \log\Big(\tan\Big (\dfrac{\pi}{3}-x\Big)\Big)dx$

$\displaystyle I=\int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx+\int_{\tfrac{\pi}{3}}^{\tfrac{5\pi}{12}} \log(\tan(x))dx+\int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{3}} \log(\tan(x))dx$

$\displaystyle I=\int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx+\int_{\tfrac{\pi}{4}}^{\tfrac{5\pi}{12}} \log(\tan(x))dx$

$\displaystyle I=\int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx-\int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{12}} \log\Big(\tan\Big (\dfrac{\pi}{2}-x\Big)\Big)dx$

$\tan\Big (\dfrac{\pi}{2}-x\Big)=\dfrac{1}{\tan(x)}$

So: $~\displaystyle I=\int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx+\int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{12}}\log(\tan(x))dx$

$\displaystyle I=2\int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx-\int_0^{\tfrac{\pi}{4}} \log(\tan(x))dx$

$\displaystyle I=2\int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx-3\int_0^{\tfrac{\pi}{12}} \log(\tan(3x))dx$

$\displaystyle I=2\int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx-3I$

$\displaystyle 2I=\int_0^{\tfrac{\pi}{12}}\log(\tan(x))dx$

Now perform change of variable $u=3x$ in the left member of $(1)$:

$\displaystyle 2\int_0^{\tfrac{\pi}{12}} \log(\tan(3x))dx=\dfrac{2}{3} \int_0^{\tfrac{\pi}{4}} \log(\tan(x))dx$

Since $~\displaystyle G=-\int_0^{\tfrac{\pi}{4}} \log(\tan(x))dx~$ then $~\displaystyle \int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx=-\dfrac{2}{3}G$.

$($Proof found in: Representations of Catalan's constant, David Bradley, $2001)$.

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  • $\begingroup$ Beautiful! Thanks a lot FDP! :) $\endgroup$ – Pranav Arora Oct 23 '14 at 20:29
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Using the Fourier series of $\ln(\tan{x})$, \begin{align} &\int^\frac{\pi}{12}_0\ln(\tan{x})\ {\rm d}x\\ =&-2\sum^\infty_{n=0}\frac{1}{2n+1}\int^\frac{\pi}{12}_0\cos\Big{[}(4n+2)x\Big{]}\ {\rm d}x\\ =&-\sum^\infty_{n=0}\frac{\sin\Big[(2n+1)\tfrac{\pi}{6}\Big{]}}{(2n+1)^2}\\ =&\color{#E2062C}{-\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+1)^2}}\color{#6F00FF}{-\sum^\infty_{n=0}\frac{1}{(12n+3)^2}}-\color{#E2062C}{\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+5)^2}}\\ &\color{#E2062C}{+\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+7)^2}}\color{#6F00FF}{+\sum^\infty_{n=0}\frac{1}{(12n+9)^2}}\color{#E2062C}{+\frac{1}{2}\sum^\infty_{n=0}\frac{1}{(12n+11)^2}}\\ =&\color{#6F00FF}{-\frac{1}{9}\underbrace{\sum^\infty_{n=0}\left[\frac{1}{(4n+1)^2}-\frac{1}{(4n+3)^2}\right]}_{G}}\color{#E2062C}{-\frac{1}{2}G-\frac{1}{2}\underbrace{\sum^\infty_{n=0}\left[\frac{1}{(12n+3)^2}-\frac{1}{(12n+9)^2}\right]}_{\frac{1}{9}G}}\\ =&\left(-\frac{1}{9}-\frac{1}{2}-\frac{1}{18}\right)G=\large{-\frac{2}{3}G} \end{align}


Things could be made clearer if we explicitly write out the terms of the sums. For the red sums, \begin{align} &-\frac{1}{2}\left(\frac{1}{1^2}+\frac{1}{5^2}-\frac{1}{7^2}-\frac{1}{11^2}+\cdots\right)\\ =&-\frac{1}{2}\left(\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}+\frac{1}{9^2}-\frac{1}{11^2}+\cdots\right)-\frac{1}{2}\left(\frac{1}{3^2}-\frac{1}{9^2}+\frac{1}{15^2}-\cdots\right)\\ =&-\frac{1}{2}G-\frac{1}{2}\cdot\frac{1}{9}\left(\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{5^2}-\cdots\right)=-\frac{5}{9}G \end{align}

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    $\begingroup$ Beautiful method! $\endgroup$ – user111187 Oct 21 '14 at 7:37
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    $\begingroup$ @user111187 Thank you. $\endgroup$ – M.N.C.E. Oct 21 '14 at 8:00
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    $\begingroup$ Thank you MNCE! Great method. Can you post some links explaining how those sums equal $G$ and $G/9$? Thanks! $\endgroup$ – Pranav Arora Oct 22 '14 at 9:46
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    $\begingroup$ @PranavArora As requested, I have added more detail. I hope my explanation helps. Thanks. $\endgroup$ – M.N.C.E. Oct 22 '14 at 11:50
  • $\begingroup$ Thank you once again MNCE! :) $\endgroup$ – Pranav Arora Oct 22 '14 at 13:53
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$\qquad\qquad\qquad\qquad$ Hello, there! Cleo just asked me to post this:

$$\int_0^\tfrac\pi{12}\ln(\tan x)~dx=-\dfrac23\cdot\text{Catalan}$$

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  • $\begingroup$ This agrees with a formula in Gradshteyn and Rhyzhik combined with a WolframAlpha evaluation. (Which is not the same as me knowing how to prove it...) $\endgroup$ – Semiclassical Oct 20 '14 at 19:58
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    $\begingroup$ I conjectured that from the numerical value provided by Wolfram Alpha. Any ideas about proving it? $\endgroup$ – Pranav Arora Oct 20 '14 at 19:58
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    $\begingroup$ +1, this answer numerically agrees. If so, then @PranavArora must prove $$\int_{-\pi/12}^{\pi/12}\ln \tan\left(\frac{\pi}{3}+x\right)\,dx=\frac{G}{3}$$ $\endgroup$ – Venus Oct 20 '14 at 20:01
  • $\begingroup$ @Venus: Yes but it is not any easier than the original integral. $\endgroup$ – Pranav Arora Oct 20 '14 at 20:04
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An alternative "elementary" method.

Consider,

\begin{align*} K&=\int_0^1 \frac{\arctan\left(\frac{x}{1-x^2}\right)}{x}\,dx\\ \end{align*} Perform the change of variable $x=\tan\left(\frac{t}{2}\right) $, \begin{align*} K&=\int_0^{\frac{\pi}{2}} \frac{\arctan\left(\frac{1}{2}\tan t\right)}{\sin t}\,dt \end{align*} Défine the function $H$ on $\left[\frac{1}{2};1\right]$ to be, \begin{align*}H(a)&=\int_0^{\frac{\pi}{2}} \frac{\arctan\left(a\tan t\right)}{\sin t}\,dt\end{align*} Observe that $K=H\left(\dfrac{1}{2}\right)$ and, \begin{align*}H(1)&=\int_0^{\frac{\pi}{2}} \frac{t}{\sin t}\,dt\\ &=\Big[t\ln\left(\tan\left(\frac{t}{2} \right)\right)\Big]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}\ln\left(\tan\left(\frac{t}{2} \right)\right)\,dt\\ &=-\int_0^{\frac{\pi}{2}}\ln\left(\tan\left(\frac{t}{2} \right)\right)\,dt\\ \end{align*} Perform the change of variable $x=\dfrac{t}{2}$, \begin{align*}H(1)&=-2\int_0^{\frac{\pi}{4}}\ln\left(\tan\left(t \right)\right)\,dt\\ &=2\text{G} \end{align*} \begin{align*}H^\prime (a)&=\int_0^{\frac{\pi}{2}} \frac{\cos x}{1-(1-a^2)\sin^2 x}\,dt\\ &=\left[\frac{1}{2\sqrt{1-a^2}}\ln\left(\frac{1+\sin(x)\sqrt{1-a^2}}{1-\sin(x)\sqrt{1-a^2}}\right)\right]_0^{\frac{\pi}{2}}\\ &=\frac{1}{2\sqrt{1-a^2}}\ln\left(\frac{1+\sqrt{1-a^2}}{1-\sqrt{1-a^2}}\right) \end{align*}
Therefore, \begin{align*}H(1)-H\left(\frac{1}{2}\right)&=\int_{\frac{1}{2}}^1 \frac{1}{2\sqrt{1-a^2}}\ln\left(\frac{1+\sqrt{1-a^2}}{1-\sqrt{1-a^2}}\right)\,da\end{align*} Perform the change of variable $y=\arctan\left(\sqrt{\dfrac{1+\sqrt{1-a^2}}{1-\sqrt{1-a^2}}}\right)$ \begin{align*}H(1)-H\left(\frac{1}{2}\right)&=-2\int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \ln\left(\tan y\right)\,dy\\ &=-2\int_0^{\frac{\pi}{4}} \ln\left(\tan y\right)\,dy+2\int_0^{\frac{\pi}{12}} \ln\left(\tan y\right)\,dy \end{align*} But, it is well known that, \begin{align*}\int_0^{\frac{\pi}{4}} \ln\left(\tan y\right)\,dy=-\text{G}\\\end{align*}

Thus,

\begin{align*}\int_0^{\frac{\pi}{12}} \ln\left(\tan y\right)\,dy=-\frac{1}{2}K\\\end{align*}

On the other hand,

\begin{align}\int_0^1 \frac{\arctan\left( \frac{x}{1-x^2}\right)}{x}\,dx-\int_0^1 \frac{\arctan x}{x}\,dx=\int_0^1 \frac{\arctan \left(x^3\right)}{x}\,dx\end{align}

In the latter integral perform the change of variable $\displaystyle y=x^3$,

\begin{align}\int_0^1 \frac{\arctan\left( \frac{x}{1-x^2}\right)}{x}\,dx-\int_0^1 \frac{\arctan x}{x}\,dx=\frac{1}{3}\int_0^1 \frac{\arctan x}{x}\,dx\end{align}

Therefore,

\begin{align}\int_0^1 \frac{\arctan\left( \frac{x}{1-x^2}\right)}{x}\,dx&=\frac{1}{3}\int_0^1 \frac{\arctan x}{x}\,dx+\int_0^1 \frac{\arctan x}{x}\,dx\\ &=\frac{4}{3}\int_0^1 \frac{\arctan x}{x}\,dx\\ &=\frac{4}{3}\text{G} \end{align}

Thus,

\begin{align*}\int_0^{\frac{\pi}{12}} \ln\left(\tan y\right)\,dy&=-\frac{1}{2}\times \frac{4}{3}\text{G} \\ &=\boxed{-\frac{2}{3}\text{G}} \end{align*}

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Rewriting the Lobachevsky functions in terms of dilogarithms, we get $$\mathcal{I}=-\frac12\Im\left[\operatorname{Li}_2\left(e^{\pi i/6}\right)+\operatorname{Li}_2\left(e^{5\pi i/6}\right)\right]=\frac12\color{blue}{\Im\left[\operatorname{Li}_2\left(e^{-\pi i/6}\right)-\operatorname{Li}_2\left(e^{5\pi i/6}\right)\right]}=-\frac23\mathbf{G},$$ where the blue expression was calculated in this answer using the triplication formula for $\operatorname{Li}_2(z)$.

Explanation: The basic building block is the integral $$\int_0^{\pi\alpha}\ln\left(2\sin x\right)dx=-\frac12\Im\operatorname{Li}_2\left(e^{2\pi i\alpha}\right),\qquad \alpha\in\left[0,1/{2}\right].$$ Writing $\ln\tan x=\ln\left(2\sin x\right)-\ln\left(2\sin (\frac{\pi}{2}-x)\right)$ and using that $\Im\operatorname{Li}_2(-1)=0$ reduces the integral to the above.

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  • $\begingroup$ Can you please show the steps to reach the final expression in terms of dilogs? I figured I would have to use the formula you linked. Since I am not very experienced with polylogs, I am finding it difficult to reach that expression from the original integral. It would be of great help to me if you could post the complete solution. Thanks a lot! $\endgroup$ – Pranav Arora Oct 21 '14 at 6:17
  • $\begingroup$ @PranavArora I've added a brief explanation. $\endgroup$ – Start wearing purple Oct 21 '14 at 13:08
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Hint:

Shifting by $u=x+\frac{\pi}{12}$,

$$\int_{-\pi/12}^{\pi/12}\ln \tan\left(\frac{\pi}{3}+x\right)\,dx=\int_{0}^{\pi/6}\ln \tan\left(\frac{\pi}{4}+u\right)\,du.$$

The integrand can be expressed via trigonometric series as:

$$\frac12\ln\tan{\left(\frac{\pi}{4}+\frac{x}{2}\right)}=\sum_{k-1}^{\infty}(-1)^{k-1}\frac{\sin{\left[(2k-1)x\right]}}{2k-1}$$

$$\implies \ln\tan{\left(\frac{\pi}{4}+u\right)}=2\sum_{k-1}^{\infty}(-1)^{k-1}\frac{\sin{\left[2(2k-1)u\right]}}{2k-1}.$$

Then,

$$\begin{align} \int_{0}^{\pi/6}\ln \tan\left(\frac{\pi}{4}+u\right)\,du &=2\int_{0}^{\pi/6}\sum_{k-1}^{\infty}(-1)^{k-1}\frac{\sin{\left[2(2k-1)u\right]}}{2k-1}\,du\\ &=2\sum_{k-1}^{\infty}\frac{(-1)^{k-1}}{2k-1}\int_{0}^{\pi/6}\sin{\left[2(2k-1)u\right]}\,du\\ &=2\sum_{k-1}^{\infty}\frac{(-1)^{k-1}}{2k-1}\cdot\frac{\cos^2{\left(\frac{\pi}{3}(k+1)\right)}}{2k-1}\\ &=2\sum_{k-1}^{\infty}\frac{(-1)^{k-1}}{(2k-1)^2}\cos^2{\left(\frac{\pi}{3}(k+1)\right)}.\\ \end{align}$$

Sums of this sort can readily be re-expressed as components of dilogarithms.

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Not an answer, just a cool series (and infinite product) representation

We will work with $$\mathrm{L}(\phi):=\int_0^\phi \log\sin x\,\mathrm dx,\qquad \phi\in (0,\pi)$$ Before we evaluate the integral, we take a look at how it relates to your integral. We define $$\begin{align} \mathrm{T}(\phi)=&\int_0^\phi\log\tan x\,\mathrm dx\\ =&\int_0^\phi\log\sin x\,\mathrm dx-\int_0^\phi\log\cos x\,\mathrm dx\\ =&\mathrm{L}(\phi)-\int_0^\phi\log\sin(x+\pi/2)\,\mathrm dx\\ =&\mathrm{L}(\phi)-\int_{\pi/2}^{\phi+\pi/2}\log\sin x\,\mathrm dx\\ =&\mathrm{L}(\phi)-\mathrm{L}(\phi+\pi/2)+\mathrm{L}(\pi/2)\\ =&\mathrm{L}(\phi)-\mathrm{L}(\phi+\pi/2)-\frac\pi2\log2 \end{align}$$

Recall that $$\sin x=x\prod_{n\geq1}\left(1-\frac{x^2}{\pi^2 n^2}\right)$$ Applying $\log$ on both sides, $$\log\sin x=\log x+\sum_{n\geq1}\log\left(1-\frac{x^2}{\pi^2 n^2}\right)$$ Then integrate over $[0,\phi]$: $$\mathrm{L}(\phi)=\phi(\log\phi-1)+\sum_{n\geq1}\phi\left[\log\frac{\pi^2n^2-\phi^2}{\pi^2n^2}-2\right]+\pi n\log\frac{\pi n+\phi}{\pi n-\phi}$$ So your integral is given by $\mathrm{T}(\pi/12)=\mathrm{L}(\pi/12)-\mathrm{L}(7\pi/12)-\frac\pi2\log2$ which boils down to the series $$\begin{align} \mathrm{T}(\pi/12)=&\frac\pi2\log\frac{6e}\pi-\frac{7\pi}{12}\log7\\&+\pi\sum_{n\geq1}\frac1{12}\left[\log\frac{144n^2-1}{144n^2e^2}+7\log\frac{144n^2e^2}{144n^2-49}\right]+n\log\frac{(12n+1)(12n-7)}{(12n-1)(12n+7)}\end{align}$$ Which can be simplified more if you so desire. If you combine all the $\log$ terms, you can use $$\log\prod_{i}a_i=\sum_{i}\log a_i$$ To turn the series into an infinite product.

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