Solve the integral $S_k = (-1)^k \int_0^1 (\log(\sin \pi x))^k dx$ My nephew asked me this, so I suggested him to sign up here. But anyways this question I was
trying to solve myself. I got part of the solution.  Let me know the rest.
$(1)$
Solve the integral defined as 
$\displaystyle{S_k =  (-1)^k \int_0^1  (\log(\sin \pi x))^k dx}$ and show
$(2)$
$\displaystyle{S_k = \frac{(-1)^k}{\sqrt\pi 2^k} \frac{d^k}{d\alpha^k} \frac{\Gamma(\alpha+\frac{1}{2})}{\Gamma(\alpha+1)}}$
with $\alpha=0$.
$(3)$
Show that 
$\displaystyle{S_4 = \frac{19 \pi^4}{240}+\frac{1}{2} \pi^2 \log^2 2 + \log^4 2 + 6 \log 2 \, \zeta(3)}$
$(4)$
Show the following:
$\displaystyle{\int_0^1 \log \log \left(\frac{1}{x}\right) \frac{dx}{1+x^2} = \frac{\pi}{2}\log \left(\sqrt{2\pi} \Gamma\left(\frac{3}{4}\right) / \Gamma\left(\frac{1}{4}\right)\right)}$
For $(3)$, if I substitute $y=\pi x$, I can transform this to a well known log-sine function as described
here at Wolfram, and so showing 
$\displaystyle{S_4 = \frac{19 \pi^4}{240}+\frac{1}{2} \pi^2 \log^2 2 + \log^4 2 + 6 \log 2 \, \zeta(3)}$
is not hard. 
I would like suggestions to read or partial solutions (not complete solutions).
(NOTE: I also noted that someone have asked a similar question for $k=2$ here)
 A: Problem (2): Use generating series.  Consider the function $$f(z)=\sum_{k=0}^{\infty}S_{k}\frac{z^{k}}{k!}.$$ Then $$f(z)=\int_0^1 \left(\sum_{k=0}^\infty (-1)^k \log^k(\sin(\pi x)) z^k \right)dx=\int_{0}^{1}\frac{1}{\sin\left(\pi x\right)^{z}}dx=\frac{\Gamma\left(\frac{1-z}{2}\right)}{\sqrt{\pi}\Gamma\left(1-\frac{z}{2}\right)}.$$  The last equality follows from an identity of the Beta Function and then applying the Duplication Formula. From here, differentiating and plugging in $z=0$ allows you to conclude (2). 
(You said you wanted few details)
A: As for the the problem (2) we have
$$
\begin{align}
S_k&=(-1)^k \int\limits_0^1\log^k(\sin\pi x)dx\\
&=\frac{(-1)^k}{\pi} \int\limits_0^\pi\log^k(\sin y)dy\\
&=\frac{(-1)^k}{\pi} \int\limits_0^{\pi/2}\log^k(\sin y)dy+
\frac{(-1)^k}{\pi} \int\limits_{\pi/2}^\pi\log^k(\sin y)dy\\
&=\frac{(-1)^k}{\pi} \int\limits_0^{\pi/2}\log^k(\sin y)dy+
\frac{(-1)^k}{\pi} \int\limits_{0}^{\pi/2}\log^k(\sin y)dy\\
&=\frac{2(-1)^k}{\pi} \int\limits_0^{\pi/2}\log^k(\sin y)dy\\
&=\frac{2(-1)^k}{\pi} \int\limits_0^{1}\frac{\log^k(t)}{\sqrt{1-t^2}}dt
\end{align}
$$
Now following Sasha's idea consider integral
$$
I(\alpha)=
\int\limits_0^1\frac{t^{2\alpha}}{\sqrt{1-t^2}}dt=
\frac{1}{2}B\left(\frac{1}{2},\frac{2\alpha+1}{2}\right)=
\frac{\sqrt{\pi}}{2}\frac{\Gamma\left(\frac{2\alpha+1}{2}\right)}{\Gamma\left(\frac{2\alpha+2}{2}\right)}
$$
Note that
$$
I^{(k)}(\alpha)=\int\limits_0^1\frac{t^{2\alpha}2^k\log^k(t)}{\sqrt{1-t^2}}dt
$$
so
$$
S_k=
\frac{2(-1)^k}{\pi} \int\limits_0^{1}\frac{\log^k(t)}{\sqrt{1-t^2}}dt=
\frac{2(-1)^k}{\pi} \frac{1}{2^k}I^{(k)}(0)=
\frac{(-1)^k}{2^k\sqrt{\pi}}\frac{d^k}{d\alpha^k}\left(\frac{\Gamma\left(\alpha+\frac{1}{2}\right)}{\Gamma\left(\alpha+1\right)}\right)_{\alpha=0}
$$
