TITLE: Evaluation of $\int_0^{\pi/3} \ln^2\left(\frac{\sin x }{\sin (x+\pi/3)}\right)\,\mathrm{d}x$
QUESTION [46 upvotes]: I plan to evaluate
$$\int_0^{\pi/3} \ln^2\left(\frac{\sin x }{\sin (x+\pi/3)}\right)\, \mathrm{d}x$$
and I need a starting point for both real and complex methods. Thanks !
Sis.

REPLY [10 votes]: Set $t=\dfrac{\sin x}{\sin\left(x+\frac\pi3\right)}$, then
$$
\int_0^{\Large\frac\pi3}\ln^2\left[\dfrac{\sin x}{\sin\left(x+\frac\pi3\right)}\right]\ dx=\frac{\sqrt3}2\int_0^1\frac{\ln^2t}{1-t+t^2}\ dt.\tag1
$$

Method $1$ :
\begin{align}
\int_0^1 \dfrac{\ln^2 t }{1-t+t^2}\ dt&= \int_0^1 \dfrac{1+t}{1+t^3} \ln^2t\ dt\\
& = \int_0^1\sum_{k=0}^{\infty}(-1)^k t^{3k}\  (1+t) \ln^2t\ dt\\
&=\sum_{k=0}^{\infty}(-1)^k\left(\int_0^1 t^{3k}\ln^2t\ dt+\int_0^1 t^{3k+1}\ln^2t\ dt\right).\tag2
\end{align}
Using
$$
\int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}, \qquad\text{for }\  n=0,1,2,\ldots\tag3
$$
then $(2)$ turns out to be
\begin{align}
\int_0^1 \dfrac{\ln^2 t }{1-t+t^2}\ dt&= \sum_{k=0}^{\infty}(-1)^k\left[\frac{2}{(3k+1)^{3}}+\frac{2}{(3k+2)^{3}}\right]\\
&= \frac2{3^3}\sum_{k=0}^{\infty}(-1)^k\left[\frac{1}{\left(k+\frac13\right)^{3}}+\frac{1}{\left(k+\frac23\right)^{3}}\right].\tag4
\end{align}
Consider the alternating series in term of polygamma function
$$
\sum_{k=0}^\infty\frac{(-1)^{k}}{(z+k)^{n+1}}=\frac1{(-2)^{n+1}n!}\left[\psi_n\left(\frac{z}{2}\right)-\psi_n\left(\frac{z+1}{2}\right)\right]\quad;\quad\text{for}\ n\in\mathbb{Z}_+\tag5
$$
and the reflection formula of polygamma function
$$
\psi_n(1-z)+(-1)^{n+1}\psi_n(z)=(-1)^{n}\pi\frac{d^n}{dz^n}\cot(\pi z).\tag6
$$
Using $(5)$ and $(6)$, then $(4)$ becomes
\begin{align}
\int_0^1 \dfrac{\ln^2 x }{1+x+x^2}\ dx&= \frac2{3^3}\cdot\frac{1}{(-2)^3\cdot2!}\bigg[\psi_2\left(\frac16\right)-\psi_2\left(\frac23\right)+\psi_2\left(\frac13\right)-\psi_2\left(\frac56\right)\bigg]\\
&= -\frac1{6^3}\bigg[\psi_2\left(\frac16\right)-\psi_2\left(\frac56\right)+\psi_2\left(\frac13\right)-\psi_2\left(\frac23\right)\bigg]\\
&= -\frac1{216}\left[\pi\left.\frac{d^2}{dz^2}\cot(\pi z)\right|_{z=\frac56}+\pi\left.\frac{d^2}{dz^2}\cot(\pi z)\right|_{z=\frac23}\right]\\
&=-\frac1{216}\left[-8\sqrt3\pi^3-\frac{8\sqrt3}{9}\pi^3\right]\\
&=\frac{10\sqrt3}{243}\pi^3.
\end{align}
Thus

$$
\int_0^{\Large\frac\pi3}\ln^2\left[\dfrac{\sin x}{\sin\left(x+\frac\pi3\right)}\right]\ dx=\frac{\sqrt3}2\int_0^1\frac{\ln^2t}{1-t+t^2}\ dt=\large\color{blue}{\frac{5}{81}\pi^3}.
$$


Method $2$ :
Set $t=\frac1x$, from $(1)$ we obtain
$$
\int_0^{\Large\frac\pi3}\ln^2\left[\dfrac{\sin x}{\sin\left(x+\frac\pi3\right)}\right]\ dx=\frac{\sqrt3}2\int_0^1\frac{\ln^2t}{1-t+t^2}\ dt=\frac{\sqrt3}4\int_0^\infty\frac{\ln^2x}{1-x+x^2}\ dx.
$$
Consider
$$
\int_0^\infty \frac{x^{\alpha}\ dx}{1+2x\cos\beta +x^{2}}=\frac{\pi\sin (\alpha\beta)}{\sin (\alpha\pi)\sin \beta }.\tag7
$$
Differentiating $(7)$ with respect to $\alpha$ twice, setting $\alpha=0$ and $\beta=\frac23\pi$, we will obtain
\begin{align}
\int_0^\infty \frac{\partial^2}{\partial\alpha^2}\left[\frac{x^{\alpha}\ dx}{1+2x\cos\beta +x^{2}}\right]_{\alpha=0\ ;\ \beta=\frac23\pi}&=\frac{\partial^2}{\partial\alpha^2}\left[\frac{\pi\sin (\alpha\beta)}{\sin (\alpha\pi)\sin \beta }\right]_{\alpha=0\ ;\ \beta=\frac23\pi}\\
\int_0^\infty \frac{\ln^2x}{1-x +x^{2}}\ dx&=\frac{20}{81\sqrt3}\pi^3
\end{align}
Thus

$$
\int_0^{\Large\frac\pi3}\ln^2\left[\dfrac{\sin x}{\sin\left(x+\frac\pi3\right)}\right]\ dx=\frac{\sqrt3}4\int_0^\infty\frac{\ln^2x}{1-x+x^2}\ dx=\large\color{blue}{\frac{5}{81}\pi^3}.
$$