7
$\begingroup$

I posted a similar question with a bad response, so I am retrying with hopes of better knowledge. The fourier series is in the form:

$$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n\cos(nx) + \sum_{n=1}^{\infty} b_n\sin(nx)$$

Where:

$$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\cos(nx) dx$$ $$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\sin(nx) dx$$

The problem is computing the coefficients. The goal of trying to derive the series is to solve:

$$\int_{0}^{\pi} \log(\sin(x)) dx$$

$\displaystyle a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} \log(\sin(x))\cos(nx) dx$

Which is very difficult to compute.

What can be done?

The series representation is:

$$\log(\sin(x)) = -\log(2) - \sum_{n=1}^{\infty} \frac{\cos(2kx)}{k}$$

$\endgroup$
  • 1
    $\begingroup$ Notice that $a_n$ is not so difficult to compute, if one performs a round of integration by parts, bringing the logarithm into a cotangent and the cosine into a sine. $\endgroup$ – Jack D'Aurizio Dec 16 '14 at 12:45
  • 1
    $\begingroup$ The only slightly tricky part is to compute $a_0$, but there is a nice proof through Riemann sums. I will be glad to show it, as soon as I have enough time (quite busy at the moment). $\endgroup$ – Jack D'Aurizio Dec 16 '14 at 12:47
  • $\begingroup$ @JackD'Aurizio, thank you for commenting. So this wont require finding the antiderivative of $\log(\sin(x))$? That is good. Again thank you Jack! $\endgroup$ – Amad27 Dec 16 '14 at 12:48
  • $\begingroup$ @JackD'Aurizio, no problem, whenever you have time. I would like to see it. Thanks once again, I will be working on this, and post an update later. Glad to hear from you! =) $\endgroup$ – Amad27 Dec 16 '14 at 12:48
  • 1
    $\begingroup$ exactly, integrating by parts we can just replace $\log\sin x$ with $\cot x$. After that, we just need Briggs' formulas to compute $a_k$ for $k\geq 1$. $\endgroup$ – Jack D'Aurizio Dec 16 '14 at 12:49
8
$\begingroup$

Ok, here I am. The goal is to compute the Fourier series of $g(x)=\log\sin x$ over $[0,\pi]$, or the Fourier series of $h(x)=\log\sin\frac{x}{2}$ over $[0,2\pi]$, or the Fourier series of $f(x)=\log\cos\frac{x}{2}$ over $[-\pi,\pi]$. Since $f(x)$ is an even function, we have to compute: $$ a_n = \frac{1}{\pi}\int_{-\pi}^{+\pi}\cos(nx)\log\cos\frac{x}{2}\,dx = \frac{2}{\pi}\int_{0}^{\pi}\cos(nx)\log\cos\frac{x}{2}\,dx$$ for any $n\geq 1$ to be able to state: $$ f(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)\,dx + \sum_{n\geq 1}a_n\cos(nx)=\frac{a_0}{2}+\sum_{n\geq 1}a_n\cos(nx).$$ for any $x\in(-\pi,\pi)$. Integration by parts gives: $$ a_n = \frac{2}{\pi}\left(\frac{1}{n}\left.\sin(nx)\log\cos\frac{x}{2}\right|_{0}^{\pi}+\frac{1}{2n}\int_{0}^{\pi}\sin(nx)\tan\frac{x}{2}\,dx\right)$$ or just: $$ a_n = \frac{1}{\pi n}\int_{0}^{\pi}\frac{\sin(nx)\sin(x/2)}{\cos(x/2)}\,dx = \frac{2}{\pi n}\int_{0}^{\pi/2}\frac{\sin(2nx)\sin x}{\cos x}\,dx.$$ Since $\cos((2n+1)x)=2\cos x\cos(2nx)-\cos((2n-1)x)$, we have: $$ a_n = \frac{1}{\pi n}\int_{0}^{\pi/2}\sum_{k=1}^{n}\cos((2k-1)x)\,dx =\frac{(-1)^{n+1}}{n}.$$ This gives: $$\log\cos\frac{x}{2}=\frac{a_0}{2}+\sum_{n\geq 1}\frac{(-1)^{n+1}}{n}\cos(nx)$$ for any $x\in(-\pi,\pi)$. In order to find $a_0$, we can simply match $f(0)=0$ with the series on the right hand side. Since: $$\sum_{n\geq 1}\frac{(-1)^{n+1}}{n}=\int_{0}^{1}\frac{dx}{1+x}=\log 2,$$ we have: $$\log\cos\frac{x}{2}=-\log 2+\sum_{n\geq 1}\frac{(-1)^{n+1}}{n}\cos(nx)\qquad\forall x\in(-\pi,\pi),\tag{1}$$ and by translating the variable: $$\log\sin\frac{x}{2}=-\log 2-\sum_{n\geq 1}\frac{1}{n}\cos(nx)\qquad\forall x\in(0,2\pi),\tag{2}$$ $$\log\sin x = -\log 2-\sum_{n\geq 1}\frac{1}{n}\cos(2nx)\qquad\forall x\in(0,\pi),\tag{3}$$ as wanted.


A little addendum, since I think it is worth mentioning the following technique. Starting with a celebrated identity: $$\prod_{k=1}^{n-1}\sin\frac{\pi k}{n}=\frac{2n}{2^n}$$ and noticing that $\log\sin x$ is a Riemann integrable function over $(0,\pi)$, we have: $$\int_{0}^{\pi}\log\sin x\,dx = \lim_{n\to +\infty}\frac{\pi}{n}\sum_{k=1}^{n-1}\log\sin\frac{\pi k}{n}=\lim_{n\to +\infty}\frac{\pi}{n}\log\frac{2n}{2^n}=\color{red}{-\pi\log 2}.$$

$\endgroup$
  • 1
    $\begingroup$ Thank you Jack! This is absolutely fantastic. I am currently away from home, I have a few questions, which if you dont mind I will ask when I get back home. Thank you once again Jack, and I apologize for what I did previously. you are an amazing person! $\endgroup$ – Amad27 Dec 16 '14 at 17:41
5
$\begingroup$

Use equation $\text{(1d)}$ below and the fact that $$ \int_0^\pi\cos(2kx)\,\mathrm{d}x=0 $$ to get $$ \int_0^\pi\log(\sin(x))\,\mathrm{d}x=-\pi\log(2) $$


$$ \begin{align} \log(\sin(x)) &=\log\left(\frac{e^{ix}-e^{-ix}}{2i}\right)\tag{1a}\\ &=+ix-\log(2)-i\pi/2+\log(1-e^{-2ix})\tag{1b}\\ &=-ix-\log(2)+i\pi/2+\log(1-e^{+2ix})\tag{1c}\\ &=-\log(2)-\sum_{k=1}^\infty\frac{\cos(2kx)}{k}\tag{1d} \end{align} $$ Explanation:
$\text{(1a)}$: $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$
$\text{(1b)}$: bring $e^{+ix}$ and $+\frac1{2i}$ out of the log
$\text{(1c)}$: bring $e^{-ix}$ and $-\frac1{2i}$ out of the log
$\text{(1d)}$: use $\log(1-x)=-\sum\limits_{k=1}^\infty\frac{x^k}k$ while averaging $\text{(1b)}$ and $\text{(1c)}$


$$ \begin{align} \log(\cos(x)) &=\log\left(\frac{e^{ix}+e^{-ix}}2\right)\tag{2a}\\ &=+ix-\log(2)+\log(1+e^{-2ix})\tag{2b}\\ &=-ix-\log(2)+\log(1+e^{+2ix})\tag{2c}\\ &=-\log(2)+\sum_{k=1}^\infty(-1)^{k-1}\frac{\cos(2kx)}{k}\tag{2d} \end{align} $$ Explanation:
$\text{(2a)}$: $\cos(x)=\frac{e^{ix}+e^{-ix}}2$
$\text{(2b)}$: bring $e^{+ix}$ and $\frac12$ out of the log
$\text{(2c)}$: bring $e^{-ix}$ and $\frac12$ out of the log
$\text{(2d)}$: use $\log(1+x)=\sum\limits_{k=1}^\infty(-1)^{k-1}\frac{x^k}k$ while averaging $\text{(2b)}$ and $\text{(2c)}$

$\endgroup$
  • $\begingroup$ I just ran across this answer in which I prove equation $\text{(1d)}$ in a different way. $\endgroup$ – robjohn Dec 19 '14 at 10:19
  • $\begingroup$ Hi Rob. Happy Holidays. Whenever I see one write $\log (z_1z_2)=\log(z_1)+\log(z_2)$ without specifying the branches that are assumed, I begin to wonder if the analysis is sound. Of course, here one only needs to ensure that the branches of $\log(1-e^{\pm i2x})$ are chosen to make the right-hand sides purely real. What did you have in mind? $\endgroup$ – Mark Viola Dec 1 '18 at 3:07
  • $\begingroup$ @MarkViola: I assumed the branch cut along the negative real axis and standard log along the positive real axis. For small $x$, things are within a narrow cone about the real axis and so there should be little problem using $\log(xy)=\log(x)+\log(y)$. $\endgroup$ – robjohn Dec 1 '18 at 6:40
3
$\begingroup$

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\dsc{\sum_{k\ = 0}^{n}\exp\pars{2ky\ic}} =\sum_{k\ = 0}^{n}\bracks{\exp\pars{2y\ic}}^{k} ={1 - \exp\pars{\bracks{n + 1}\bracks{2y\ic}} \over 1 - \exp\pars{2y\ic}} \\[5mm]&={\exp\pars{-y\ic} - \exp\pars{\bracks{2n + 1}y\ic}\over \exp\pars{-y\ic} - \exp\pars{y\ic}} =\dsc{\half\,\ic\,{\exp\pars{-y\ic} - \exp\pars{\bracks{2n + 1}y\ic} \over \sin\pars{y}}} \end{align}

We'll take the imaginary part in both members: \begin{align}&\sum_{k\ = 1}^{n}\sin\pars{2ky} =\half\,\cot\pars{y} - {\cos\pars{\bracks{2n + 1}y} \over 2\sin\pars{y}} \end{align}

With $\ds{0 < x \leq {\pi \over 2}}$, integrate in both members over $\ds{\pars{x,{\pi \over 2}}}$: \begin{align}&\sum_{k\ = 1}^{n} {-\cos\pars{2k\bracks{\pi/2}} + \cos\pars{2kx} \over 2k} =\left.\half\,\ln\pars{\sin\pars{y}}\right\vert_{x}^{\pi/2} -\half\int_{x}^{\pi/2}{\cos\pars{\bracks{2n + 1}y} \over \sin\pars{y}}\,\dd y \end{align}

Then, \begin{align}-\ \overbrace{\sum_{k\ = 1}^{n}{\pars{-1}^{k}\over k}} ^{\dsc{-\ln\pars{2}}} +\sum_{k\ = 1}^{n}{\cos\pars{2kx} \over k} =-\ln\pars{\sin\pars{x}} -\int_{x}^{\pi/2}{\cos\pars{\bracks{2n + 1}y} \over \sin\pars{y}}\,\dd y \end{align}

We'll take the limit $\ds{n \to\ \infty}$: \begin{align}\ln\pars{\sin\pars{x}}&= -\ln\pars{2} - \sum_{k\ = 1}^{\infty}{\cos\pars{2kx} \over k}\ -\ \underbrace{\lim_{n\ \to\ \infty} \int_{x}^{\pi/2}{\cos\pars{\bracks{2n + 1}y} \over \sin\pars{y}}\,\dd y}_{\ds{=}\ \dsc{0}} \end{align}

$$ \color{#66f}{\large\ln\pars{\sin\pars{x}}}= \color{#66f}{\large-\ln\pars{2} - \sum_{k\ = 1}^{\infty}{\cos\pars{2kx} \over k}} \,,\qquad x \in \left(\, 0,{\pi \over 2}\,\right]\tag{1} $$ Indeed $\ds{\large\it\mbox{it's still valid}}$ in $\pars{0,\pi}$ $\ds{\pars{~\color{#00f}{\sf\mbox{see below}}~}}$

When $\ds{x \in \pars{{\pi \over 2},\pi}}$: \begin{align} \ln\pars{\sin\pars{x}}&=\ln\pars{\sin\pars{\pi - x}} =-\ln\pars{2} - \sum_{k\ = 1}^{\infty}{\cos\pars{2k\bracks{\pi - x}} \over k} \\[5mm]&=-\ln\pars{2} - \sum_{k\ = 1}^{\infty}{\cos\pars{2kx} \over k} \end{align} such that $\pars{1}$ is valid in $\pars{0,\pi}$.

$\endgroup$

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.