Double integral $ \iint\ln(\sin(u-7v)) \,du\,dv$

Question

Can you please help me solve this double integral?

$$ \iint_D\ln(\sin(u-7v))\,du\,dv $$ where $$D:=\left\{(u,v)\;:\; 0\le u \le \pi \;,\; 0\le v \le \frac u7\right\} .$$

I know it's not possible to solve $\int \ln(\sin(x))$ but definite integral is?

Answer 1

By replacing the variable $v$ with $t:=u-7v$, we get $$\int_{u=0}^{\pi}\int_{v=0}^{u/7}\ln(\sin(u-7v))\,du\,dv= \frac{1}{7}\int_{t=0}^{\pi}\int_{u=t}^{\pi}\ln(\sin(t))\,du\,dt\\ =\frac{1}{7}\int_{t=0}^{\pi}\ln(\sin(t))(\pi-t)\,dt=\frac{I}{7}=\fbox{$-\frac{\pi^2\ln(2)}{14}$}$$ where $I=\int_{t=0}^{\pi}t\ln(\sin(t))\,dt=-\frac{\pi^2\ln(2)}{2}$. In fact $$I=\int_{t=0}^{\pi}\ln(\sin(t))(\pi-t)\,dt=\pi\int_{t=0}^{\pi}\ln(\sin(t))\,dt-I$$ which implies $$I=\frac{\pi}{2}\int_{t=0}^{\pi}\ln(\sin(t))\,dt=\pi\int_{t=0}^{\pi/2}\ln(\sin(t))\,dt=-\frac{\pi^2\ln(2)}{2}$$ where in the last step we used Today a student asked me $\int \ln (\sin x) \, dx.$

Answer 2

Did not it help you if rewrite it this way?

$$\int_0^\pi du \int_0^{\frac{u}{7}} \ln(\sin(u-7v))\,dv$$