2
$\begingroup$

$$ \int_{0}^{\pi}\int_{0}^{x/8}\ln\left(\,\sin\left(\,x - 8y\,\right)\,\right) \,\mathrm{d}y\,\mathrm{d}x $$ 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.

$\endgroup$
  • $\begingroup$ Numerically I got that answer only to 8 decimal digits using Maple. $\endgroup$ – i. m. soloveichik Jun 18 '17 at 14:31
3
$\begingroup$

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 $$\int_{0}^{\pi}\int_{0}^{x}\cos\left(2k\left(x-y\right)\right)dydx=\frac{1}{2k}\int_{0}^{\pi}\sin\left(2kx\right)dx=0.$$

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.