# 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.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}