# 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.

• Numerically I got that answer only to 8 decimal digits using Maple. – i. m. soloveichik Jun 18 '17 at 14:31

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 $$\int_{0}^{\pi}\int_{0}^{x}\cos\left(2k\left(x-y\right)\right)dydx=\frac{1}{2k}\int_{0}^{\pi}\sin\left(2kx\right)dx=0.$$
$$\begin{eqnarray*}I=\int_{0}^{\pi}\int_{0}^{x/8}\log\sin(x-8y)\,dy\,dx &=& \int_{0}^{\pi}\int_{0}^{1}\frac{x}{8}\log\sin(x-xz)\,dz\,dx\\&=&\frac{1}{8}\int_{0}^{\pi}\int_{0}^{1} x\log\sin(xw)\,dw\,dx\tag{1}\end{eqnarray*}$$ 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.