How do I solve this double definite integral? $$
\int_{0}^{\pi}\int_{0}^{x/8}\ln\left(\,\sin\left(\,x - 8y\,\right)\,\right)
\,\mathrm{d}y\,\mathrm{d}x
$$ 
I am pretty sure the solution
is $\displaystyle-\,\frac{\ln\left(\,2\,\right)\,\pi^{2}}{16}$. I just don't know how to get there.
I tried using the method for
solving $\int_{0}^{\pi/2}\ln\left(\sin\left(x\right)\right)
\,\mathrm{d}x = -\ln\left(2\right)\pi/2$, but I can't figure out the limits.
 A: From the Fourier series of $\log\left(\sin\left(z\right)\right)$ we have $$\int_{0}^{\pi}\int_{0}^{x/8}\log\left(\sin\left(x-8y\right)\right)dydx\stackrel{8y\rightarrow y}{=}\frac{1}{8}\int_{0}^{\pi}\int_{0}^{x}\log\left(\sin\left(x-y\right)\right)dydx$$ $$=-\frac{\log\left(2\right)}{16}\pi^{2}-\frac{1}{8}\sum_{n\geq1}\frac{1}{k}\int_{0}^{\pi}\int_{0}^{x}\cos\left(2k\left(x-y\right)\right)dydx$$ and the last integrals are quite simple to evaluate $$\int_{0}^{\pi}\int_{0}^{x}\cos\left(2k\left(x-y\right)\right)dydx=\frac{1}{2k}\int_{0}^{\pi}\sin\left(2kx\right)dx=0.$$
A: $$\begin{eqnarray*}I=\int_{0}^{\pi}\int_{0}^{x/8}\log\sin(x-8y)\,dy\,dx &=& \int_{0}^{\pi}\int_{0}^{1}\frac{x}{8}\log\sin(x-xz)\,dz\,dx\\&=&\frac{1}{8}\int_{0}^{\pi}\int_{0}^{1} x\log\sin(xw)\,dw\,dx\tag{1}\end{eqnarray*}$$
and by exploiting the Fourier series of $\log\sin$ we have:
$$\begin{eqnarray*} \int_{0}^{1}\log\sin(xw)\,dw &=& -\log(2)-\sum_{k\geq 1}\int_{0}^{1}\frac{\cos(2kxw)}{k}\,dw\\&=&-\log(2)-\color{blue}{\sum_{k\geq 1}\frac{\sin(2kx)}{2k^2 x}}\tag{2}\end{eqnarray*}$$
so by multiplying both sides of $(2)$ by $x$ and by applying $\frac{1}{8}\int_{0}^{\pi}(\ldots)\,dx$ we simply get $I=\color{red}{\large -\frac{\pi^2\log(2)}{16}}$, since the blue series does not contribute at all.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
&\int_{0}^{\pi}\int_{0}^{x/8}\ln\pars{\sin\pars{x - 8y}}\,\dd y\,\dd x
\,\,\,\stackrel{y\ \mapsto\ x/8 - y}{=}\,\,\,
\int_{0}^{\pi}\int_{0}^{x/8}\ln\pars{\sin\pars{x - 8\bracks{{x \over 8} - y}}}
\,\dd y\,\dd x
\\[5mm] = &\
\int_{0}^{\pi}\int_{0}^{x/8}\ln\pars{\sin\pars{8y}}\,\dd y\,\dd x
\,\,\,\stackrel{8y\ \mapsto\ y}{=}\,\,\,
{1 \over 8}\int_{0}^{\pi}\int_{0}^{x}\ln\pars{\sin\pars{y}}\,\dd y\,\dd x
\\[5mm] = &\
{1 \over 8}\int_{0}^{\pi}\ln\pars{\sin\pars{y}}\int_{y}^{\pi}\,\dd x\,\dd y =
{1 \over 8}\int_{0}^{\pi}\ln\pars{\sin\pars{y}}\pars{\pi - y}\,\dd y
\\[5mm] = &\
{1 \over 8}\int_{-\pi/2}^{\pi/2}\ln\pars{\cos\pars{y}}
\pars{{\pi \over 2} - y}\,\dd y =
{1 \over 8}\,\pi\int_{0}^{\pi/2}\ln\pars{\cos\pars{y}}\,\dd y
\\[5mm] = &\,\,\,
\overbrace{\left.{1 \over 8}\,\pi\,\Re\int_{\theta = 0}^{\theta = \pi/2}
\ln\pars{1 + z^{2} \over 2z}\,{\dd z \over \ic z}
\right\vert_{\ z\ =\ \exp\pars{\ic\theta}}}
^{\ds{\ln\,\,\, \mbox{is its}\ Principal\ Branch}}\ =\
\left.{1 \over 8}\,\pi\
\Im\int_{\theta = 0}^{\theta = \pi/2}
\ln\pars{1 + z^{2} \over 2z}\,{\dd z \over z}
\right\vert_{\ z\ =\ \exp\pars{\ic\theta}}
\\[1cm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim} &\
-\,{1 \over 8}\,\pi\,\Im\int_{1}^{\epsilon}
\overbrace{\ln\pars{-\,{1 - y^{2} \over 2y}\,\ic}}
^{\ds{\ln\pars{1 - y^{2} \over 2y} - {\pi \over 2}\,\ic}}\
\,{\ic\,\dd y \over \ic y} -
{1 \over 8}\,\pi\,\
\overbrace{\Im\int_{\pi/2}^{0}
\ln\pars{{1 \over 2\epsilon}\,\expo{-\ic\theta}}\,{\epsilon\expo{\ic\theta}\ic\,\dd\theta \over \epsilon\expo{\ic\theta}}}
^{\ds{\int_{\pi/2}^{0}\ln\pars{1 \over 2\epsilon}\,\dd\theta}}
\\[2mm] &\ -\
\underbrace{{1 \over 8}\,\pi\,\Im\int_{\epsilon}^{1}
\ln\pars{1 + x^{2} \over 2x}\,{\dd x \over x}}_{\ds{=\ 0}}
\\[1cm] = &\
-\,{1 \over 8}\,\pi\pars{-\,{\pi \over 2}}\ln\pars{\epsilon} -
{1 \over 8}\,\pi\pars{-\,{\pi \over 2}}
\bracks{\vphantom{\large A}-\ln\pars{2} - \ln\pars{\epsilon}}
\\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\to}\,\,\,&
\bbx{-\,{1 \over 16}\,\pi^{2}\ln\pars{2}}
\end{align}
