Krdytkyu

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.

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

Hint:

As shown in this answer,
$$
log(sin(x))=-log(2)-sum_{k=1}^inftyfrac{cos(2kx)}{k}
$$
Therefore,
$$
intlog(sin(x)),mathrm{d}x=-xlog(2)-sum_{k=1}^inftyfrac{sin(2kx)}{2k^2}
$$

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

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&bbox[10px,#ffd]{ds{%
int_{0}^{pi/3}lnpars{sinpars{x} over sinpars{x + pi/3}},dd x}} =
int_{-pi/6}^{pi/6}lnpars{sinpars{x + pi/6} over cospars{x}},dd x
\[5mm] = &
int_{-pi/6}^{pi/6}lnpars{{root{3} over 2},tanpars{x} + { 1 over 2}}
,dd x
,,,,,,stackrel{large x = arctanpars{2t - 1 over root{3}}}{=},,,
,,,
{root{3} over 2}int_{0}^{1}{lnpars{t} over t^{2} - t + 1},dd t
\[5mm] = &
{root{3} over 2}int_{0}^{1}{lnpars{t} over pars{t - r}pars{t - bar{r}}},dd tlabel{1}tag{1}
end{align}

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

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

The given integral can be approached as follows:
$$begin{eqnarray*} mathcal{I}=int_{0}^{pi/3}logleft(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}cottheta),dtheta\&=&frac{pi}{3}log 2-int_{1/sqrt{3}}^{+infty}frac{log(1+tsqrt{3})}{1+t^2},dt\&=&frac{pi}
{3}log2-int_{0}^{sqrt{3}}frac{logleft(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}{24sqrt{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_{kgeq 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_{kgeq 0}left(frac{1}{(6k+1)^2}color{red}{+}frac{1}{(6k+5)^2}right)=frac{pi^2}{9},qquad sum_{kgeq 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 $logsin(x)$, which is well-known.

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