integral $ \int_0^{\frac{\pi}{3}}\mathrm{ln}\left(\frac{\mathrm{sin}(x)}{\mathrm{sin}(x+\frac{\pi}{3})}\right)\ \mathrm{d}x$

Question

We want to evaluate

$ \displaystyle \int_0^{\frac{\pi}{3}}\mathrm{ln}\left(\frac{\mathrm{sin}(x)}{\mathrm{sin}(x+\frac{\pi}{3})}\right)\ \mathrm{d}x$.

We tried contour integration which was not helpful. Then we started trying to use symmetries, this also doesn't work.

Answer 1

Here is an approach that uses real methods only.

\begin{align*} I &= \int_0^{\pi/3} \ln \left (\frac{\sin x}{\sin \left (x + \frac{\pi}{3} \right )} \right ) \, dx\\ &= \int_0^{\pi/3} \ln (\sin x) \, dx - \int_0^{\pi/3} \ln \left (\sin \left (x + \frac{\pi}{3} \right ) \right ) \, dx. \end{align*} If in the second integral appearing on the right a substitution of $x \mapsto x - \frac{\pi}{3}$ is enforced, one has \begin{align*} I &= \int_0^{\pi/3} \ln (\sin x) \, dx - \int_{\pi/3}^{2\pi/3} \ln (\sin x) \, dx\\ &= 2 \int_0^{\pi/3} \ln (\sin x) \, dx - \int_0^{2\pi/3} \ln (\sin x) \, dx.\tag1 \end{align*}

Now consider the integral $$I(\alpha) = \int_0^\alpha \ln (\sin x) \, dx, \quad 0 < \alpha < \pi.\tag2$$ Taking advantage of the well-known identity $$\ln (\sin x) = -\ln 2 - \sum_{k = 1}^\infty \frac{\cos (2kx)}{k}, \quad 0 < x < \pi,$$ substituting this result into (2), after interchanging the summation with the integration before integrating one finds $$I(\alpha) = -\alpha \ln 2 - \frac{1}{2} \sum_{k = 1}^\infty \frac{\sin (2k \alpha)}{k^2} = -\alpha \ln 2 - \frac{1}{2} \text{Cl}_2 (\alpha).$$ Here $\text{Cl}_2 (\varphi)$ denotes the Clausen function of order two.

In terms of the Clausen function of order two the integral in (1) can be written as $$I = -\text{Cl}_2 \left (\frac{2\pi}{3} \right ) + \frac{1}{2} \text{Cl}_2 \left (\frac{4\pi}{3} \right ).\tag3$$ From the duplication formula for the Clausen function of order two, namely $$\text{Cl}_2 (2\theta) = 2 \text{Cl}_2 (\theta) - 2 \text{Cl}_2 (\pi - \theta), \quad 0 < \theta < \pi,$$ if we set $\theta = 2\pi/3$ in the above duplication formula, as $$\text{Cl}_2 \left (\frac{4\pi}{3} \right ) = 2 \text{Cl}_2 \left (\frac{2\pi}{3} \right ) - 2 \text{Cl}_2 \left (\frac{\pi}{3} \right ),$$ the expression for our integral in (3) can be expressed more simply as $$\int_0^{\pi/3} \ln \left (\frac{\sin x}{\sin \left (x + \frac{\pi}{3} \right )} \right ) \, dx = -\text{Cl}_2 \left (\frac{\pi}{3} \right ).$$

Answer 2

Hint:

As shown in this answer, $$ \log(\sin(x))=-\log(2)-\sum_{k=1}^\infty\frac{\cos(2kx)}{k} $$ Therefore, $$ \int\log(\sin(x))\,\mathrm{d}x=-x\log(2)-\sum_{k=1}^\infty\frac{\sin(2kx)}{2k^2} $$

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Applying the hint: $$ \begin{align} \int_0^{\pi/3}\log\left(\frac{\sin(x)}{\sin(x+\pi/3)}\right)\,\mathrm{d}x &=\int_0^{\pi/3}\log(\sin(x))\,\mathrm{d}x-\int_{\pi/3}^{2\pi/3}\log(\sin(x))\,\mathrm{d}x\\ &=-2\sum_{k=1}^\infty\frac{\sin(2k\pi/3)}{2k^2}+\sum_{k=1}^\infty\frac{\sin(4k\pi/3)}{2k^2}\\ &=-\sum_{k=1}^\infty\frac{\sin(2k\pi/3)}{k^2}+2\sum_{k=1}^\infty\frac{\sin(4k\pi/3)}{4k^2}\\ &=\sum_{k=1}^\infty(-1)^k\frac{\sin(2k\pi/3)}{k^2}\\ &=-\sum_{k=1}^\infty\frac{\sin(k\pi/3)}{k^2}\\ &=-\frac{\sqrt3}2\sum_{k=0}^\infty(-1)^k\left(\frac1{(3k+1)^2}+\frac1{(3k+2)^2}\right) \end{align} $$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\ds{% \int_{0}^{\pi/3}\ln\pars{\sin\pars{x} \over \sin\pars{x + \pi/3}}\,\dd x}} = \int_{-\pi/6}^{\pi/6}\ln\pars{\sin\pars{x + \pi/6} \over \cos\pars{x}}\,\dd x \\[5mm] = &\ \int_{-\pi/6}^{\pi/6}\ln\pars{{\root{3} \over 2}\,\tan\pars{x} + { 1 \over 2}} \,\dd x \,\,\,\,\,\,\stackrel{\large x\ =\ \arctan\pars{2t - 1 \over \root{3}}}{=}\,\,\, \,\,\, {\root{3} \over 2}\int_{0}^{1}{\ln\pars{t} \over t^{2} - t + 1}\,\dd t \\[5mm] = &\ {\root{3} \over 2}\int_{0}^{1}{\ln\pars{t} \over \pars{t - r}\pars{t - \bar{r}}}\,\dd t\label{1}\tag{1} \end{align}

where $\ds{r \equiv {1 \over 2} + {\root{3} \over 2}\,\ic = \exp\pars{{\pi \over 3}\,\ic}}$

\eqref{1} is reduced to \begin{align} &\bbox[10px,#ffd]{\ds{% \int_{0}^{\pi/3}\ln\pars{\sin\pars{x} \over \sin\pars{x + \pi/3}}\,\dd x}} = {\root{3} \over 2}\int_{0}^{1}\ln\pars{t} \pars{{1 \over t - r} - {1 \over t - \bar{r}}}{1 \over r - \bar{r}}\,\dd t \\[5mm] = &\ {\root{3} \over 2}\,{1 \over 2\ic\,\Im\pars{r}}\,2\ic\,\Im\int_{0}^{1}{\ln\pars{t} \over t - r}\,\dd t = -\,\Im\int_{0}^{1}{\ln\pars{t} \over r - t}\,\dd t = -\,\Im\int_{0}^{1/r}{\ln\pars{rt} \over 1 - t}\,\dd t \\[5mm] = &\ -\,\Im\int_{0}^{\large\bar{r}}{\ln\pars{1 - t} \over t}\,\dd t = \Im\int_{0}^{\large\bar{r}}\mrm{Li}_{2}'\pars{t}\,\dd t\qquad \pars{~\mrm{Li}_{s}\ \mbox{is the}\ PolyLogarithm\ Function~} \\[5mm] \implies &\ \bbx{\int_{0}^{\pi/3}\ln\pars{\sin\pars{x} \over \sin\pars{x + \pi/3}}\,\dd x = \Im\mrm{Li}_{2}\pars{\expo{-\pi\ic/3}} \approx -1.0149} \end{align}

Answer 4

Hint: $\displaystyle \int_0^{\dfrac{\pi}{3}}\mathrm{ln}\left(\frac{\mathrm{sin}(x)}{\mathrm{sin}\left(x+\frac{\pi}{3}\right)}\right)\ \mathrm{d}x=\displaystyle \int_0^{\dfrac{\pi}{3}}\mathrm{ln}({\mathrm{sin}(x)})dx-\int_0^{\dfrac{\pi}{3}} \ln\left({\mathrm{sin}\left(x+\frac{\pi}{3}\right)}\right)\ \mathrm{d}x$

Answer 5

The given integral can be approached as follows: $$\begin{eqnarray*} \mathcal{I}=\int_{0}^{\pi/3}\log\left(\frac{\sin x}{\frac{1}{2}\sin x+\frac{\sqrt{3}}{2}\cos x}\right)\,dx&=&\frac{\pi}{3}\log 2-\int_{0}^{\pi/3}\log(1+\sqrt{3}\cot\theta)\,d\theta\\&=&\frac{\pi}{3}\log 2-\int_{1/\sqrt{3}}^{+\infty}\frac{\log(1+t\sqrt{3})}{1+t^2}\,dt\\&=&\frac{\pi} {3}\log2-\int_{0}^{\sqrt{3}}\frac{\log\left(1+\frac{\sqrt{3}}{t}\right)}{1+t^2}\,dt\\&=&\frac{\pi}{3}\log2-\sqrt{3}\int_{1}^{+\infty}\frac{\log(z+1)}{3+z^2}\,dz\\(\text{trigamma})\qquad&=&-\frac{1}{24\sqrt{3}}\left(\psi'\left(\tfrac{1}{6}\right)+\psi'\left(\tfrac{1}{3}\right)-\psi'\left(\tfrac{2}{3}\right)-\psi'\left(\tfrac{5}{6}\right)\right)\end{eqnarray*} $$ from which it follows that $$\mathcal{I}=-\frac{\sqrt{3}}{2}\sum_{k\geq 0}\left(\frac{1}{(6k+1)^2}+\frac{1}{(6k+2)^2}-\frac{1}{(6k+4)^2}-\frac{1}{(6k+5)^2}\right).\tag{1}$$ We don't have a really nicer closed form, but the terms of the last series behaves like $\frac{1}{18k^3}$ for large values of $k$, ensuring a decent speed of convergence of the partial sums. Due to the reflection formulas for the trigamma function we have $$\sum_{k\geq 0}\left(\frac{1}{(6k+1)^2}\color{red}{+}\frac{1}{(6k+5)^2}\right)=\frac{\pi^2}{9},\qquad \sum_{k\geq 0}\left(\frac{1}{(6k+2)^2}\color{red}{+}\frac{1}{(6k+4)^2}\right)=\frac{\pi^2}{27} $$ which can be used to devise acceleration tricks, even if the signs pattern in the previous identity is the "wrong" one. Numerically $\mathcal{I}\approx -1.0149416$.

$(1)$ can be proved also by applying termwise integration to the Fourier series of $\log\sin(x)$, which is well-known.