For $\alpha, \beta, \gamma \in (0,1)$ satisfying $\alpha+\beta+\gamma = 1$ and
$\mu \in \mathbb{C} \setminus [1,\infty)$, define
$$
F_{\alpha\beta}(\mu) = \int_0^1\frac{dx}{x^\alpha(1-x)^\beta(1-\mu x)^\gamma}
\quad\text{ and }\quad
\Delta = \frac{\Gamma(1-\alpha)\Gamma(1-\beta)}{\Gamma(1+\gamma)}
$$
When $|\mu| < 1$, we can rewrite the integral $F_{\alpha\beta}(\mu)$ as
$$\begin{align}
F_{\alpha\beta}(\mu)
= & \int_0^1 \frac{1}{x^\alpha(1-x)^{\beta}}\left(\sum_{n=0}^{\infty}\frac{(\gamma)_n}{n!}\mu^n x^n\right) dx
= \sum_{n=0}^{\infty}\frac{(\gamma)_n}{n!}\frac{\Gamma(n+1-\alpha)\Gamma(1-\beta)}{\Gamma(n+1+\gamma)}\mu^n\\
= & \Delta\sum_{n=0}^{\infty}\frac{(\gamma)_n (1-\alpha)_n}{n!(\gamma+1)_n}\mu^n
= \Delta\gamma \sum_{n=0}^{\infty}\frac{(1-\alpha)_n}{n!(\gamma+n)}\mu^n
\end{align}$$
This implies
$$
\mu^{-\gamma} \left(\mu\frac{\partial}{\partial \mu}\right) \mu^{\gamma} F_{\alpha\beta}(\mu) =
\Delta\gamma \sum_{n=0}^{\infty}\frac{(1-\alpha)_n}{n!}\mu^n
= \Delta\gamma\frac{1}{(1-\mu)^{1-\alpha}}
$$
and hence
$$F_{\alpha\beta}(\mu)
= \Delta\gamma \mu^{-\gamma} \int_0^\mu \frac{\nu^{\gamma-1}d\nu}{(1-\nu)^{1-\alpha}}
= \Delta\gamma \int_0^1 \frac{t^{\gamma-1} dt}{(1-\mu t)^{1-\alpha}}
= \Delta \int_0^1 \frac{dt}{(1 - \mu t^{1/\gamma})^{1-\alpha}}$$
Notice if we substitute $x$ by $y = 1-x$, we have
$$F_{\alpha\beta}(\mu) = \int_0^1 \frac{dy}{y^\beta(1-y)^\alpha(1-\mu - \mu y)^{\gamma}}
= \frac{1}{(1-\mu)^\gamma} F_{\beta\alpha}(-\frac{\mu}{1-\mu})$$
Combine these two representations of $F_{\alpha\beta}(\mu)$ and let $\omega = \left(\frac{\mu}{1-\mu}\right)^{\gamma}$, we obtain
$$F_{\alpha\beta}(\mu) = \frac{\Delta}{(1-\mu)^{\gamma}}\int_0^1 \frac{dt}{( 1 + \omega^{1/\gamma} t^{1/\gamma})^{1-\beta}} = \frac{\Delta}{\mu^\gamma}\int_0^\omega \frac{dt}{(1 + t^{1/\gamma})^{1-\beta}}$$
Let $(\alpha,\beta,\gamma) = (\frac14,\frac12,\frac14)$ and $\mu = \frac{\sqrt{3}}{2}$, the identity we want to check becomes
$$\frac{\Gamma(\frac34)\Gamma(\frac12)}{\Gamma(\frac54) (\sqrt{3})^{1/4}}\int_0^\omega \frac{dt}{\sqrt{1+t^4}} \stackrel{?}{=} \frac{2\sqrt{2}}{3\sqrt[8]{3}} \pi\tag{*1}$$
Let $K(m)$ be the complete elliptic integral of the first kind associated with modulus $m$. i.e.
$$K(m) = \int_0^1 \frac{dx}{\sqrt{(1-x^2)(1-mx^2)}}$$
It is known that $\displaystyle K(\frac12) = \frac{8\pi^{3/2}}{\Gamma(-\frac14)^2}$. In term of $K(\frac12)$, it is easy to check $(*1)$ is equivalent to
$$\int_0^\omega \frac{dt}{\sqrt{1+t^4}} \stackrel{?}{=} \frac23 K(\frac12)\tag{*2}$$
To see whether this is the case, let $\varphi(u)$ be the inverse function of above integral.
More precisely, define $\varphi(u)$ by following relation:
$$u = \int_0^{\varphi(u)} \frac{dt}{\sqrt{1+t^4}}$$
Let $\psi(u)$ be $\frac{1}{\sqrt{2}}(\varphi(u) + \varphi(u)^{-1})$. It is easy to check/verify
$$
\varphi'(u)^2 = 1 + \varphi(u)^4
\implies
\psi'(u)^2 = 4 (1 - \psi(u)^2)(1 - \frac12 \psi(u)^2)
$$
Compare the ODE of $\psi(u)$ with that of a Jacobi elliptic functions with modulus $m = \frac12$, we find
$$\psi(u) = \text{sn}(2u + \text{constant} | \frac12 )\tag{*3}$$
Since we are going to deal with elliptic functions/integrals with $m = \frac12$ only,
we will simplify our notations and drop all reference to modulus, i.e
$\text{sn}(u)$ now means $\text{sn}(u|m=\frac12)$ and $K$ means $K(m = \frac12)$.
Over the complex plane, it is known that $\text{sn}(u)$ is doubly periodic with
fundamental period $4 K$ and $2i K$. It has two poles at $i K$ and $(2 + i)K$ in the fundamental domain.
When $u = 0$, we want $\varphi(u) = 0$ and hence $\psi(u) = \infty$. So the constant
in $(*3)$ has to be one of the pole. For small and positive $u$, we want $\varphi(u)$ and hence $\psi(u)$ to be positive. This fixes the constant to $i K$. i.e.
$$\psi(u) = \text{sn}(2u + iK )$$
and the condition $(*2)$ becomes whether following equality is true or not.
$$\frac{1}{\sqrt{2}} (\omega + \omega^{-1}) \stackrel{?}{=} \text{sn}( \frac43 K + i K)\tag{*4}$$
Notice $ 3( \frac43 K + i K) = 4 K + 3 i K $ is a pole of $\text{sn}(u)$. if one repeat
apply the addition formula for $\text{sn}(u+v)$
$$\text{sn}(u+v) = \frac{\text{sn}(u)\text{cn}(v)\text{dn}(v)+\text{sn}(v)\text{cn}(u)\text{dn}(u)}{1-m\,\text{sn}(u)^2 \text{sn}(v)^2}$$
One find in order for $\text{sn}(3u)$ to blow up, $\text{sn}(u)$ will be a root of
following polynomial equation:
$$3 m^2 s^8-4 m^2 s^6-4 m s^6+6 m s^4-1 = 0$$
Substitute $m = \frac12$ and $s = \frac{1}{\sqrt{2}}(t+\frac{1}{t})$ into this, the equation $\omega$ need to satisfy is given by:
$$(t^8 - 6 t^4 - 3)(3 t^8 + 6 t^4 - 1 ) = 0$$
One can check that $\omega = \sqrt[4]{\frac{\sqrt{3}}{2-\sqrt{3}}}$ is indeed a root of this polynomial. As a result, the original equality is valid.
Persistence pays off! Given the log sine integral,
$$\rm{Ls}_n\Big(\frac{\pi}3\Big) = \int_0^{\pi/3}\ln^{n-1}\big(2\sin\tfrac{\theta}{2}\big)\,d\theta$$
The case $\rm{Ls}_7\big(\frac{\pi}3\big)$ was elusive, but $\rm{Ls}_\color{red}8\big(\frac{\pi}3\big)$ was found. Hence,
$$\frac{2^{10}\cdot9\pi}{7!} \int_0^{\pi/3}\ln^7\big(2\sin\tfrac{x}{2}\big)\,dx+6^3 \sum_{n=1}^\infty \frac{1}{n^9\,\binom {2n}n}\\=-13921\zeta(9)-6^4\zeta(2)\zeta(7)-6087\zeta(3)\zeta(6)-4428\zeta(4)\zeta(5)-192\zeta^3(3) $$
though I don't know why the lower level is more elusive.
Best Answer
In general, we have $$\begin{align} \operatorname{ls}^{(1)}_k(a) := \int_0^a x \, \ln \left( \frac{\sin x}{\sin a} \right)^{k-1} \, \mathrm{d}x \\&= \sin a \int_0^1 \dfrac{\sin^{-1}( x \, \sin a)}{\sqrt{1-x^2 \sin^2 a}} \, \ln(x)^{k-1}\, \mathrm{d}x \\&= \sin a \int_0^1 \sum_{n=1}^\infty \frac{(2 x \sin a)^{2n-1}}{n \binom{2n}{n}}\, \ln(x)^{k-1}\, \mathrm{d}x \\&= (k-1)! \,(-2)^{-k-1} \sum_{n=1}^\infty \frac{(2 \sin a)^{2n}}{n^{k+1} \binom{2n}{n}} \end{align}$$ If we define $$\operatorname{ls}_k(a) := \int_0^a \ln \left( \frac{\sin x}{\sin a} \right)^{k-1} \, \mathrm{d}x,$$ we may note that your previous post establishes closed forms in term of zeta functions for $$\frac{\pi}{6} \operatorname{ls}_k\left(\frac{\pi}{6}\right)- \operatorname{ls}^{(1)}_k\left(\frac{\pi}{6}\right)$$ for $k \in {1,2,3,4,5,6,8}.$