10
$\begingroup$

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:

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

$\endgroup$
6
  • 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. $\endgroup$ 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). $\endgroup$ Dec 16 '14 at 12:47
  • 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 $k\geq 1$. $\endgroup$ Dec 16 '14 at 12:49
  • $\begingroup$ @JackD'Aurizio, it is very hard to compute the integral. I came up with: $$I = \bigg[\log(\sin(x))\sin(nx)/n\bigg]_{-\pi}^{\pi} - \frac{1}{n}\cdot \int_{-\pi}^{\pi} \frac{\sin(nx)\cos(x)}{\sin(x)} dx$$ But the integral on the RHS does not converge $\endgroup$
    – Amad27
    Dec 16 '14 at 13:10
  • $\begingroup$ look carefully. $\frac{\sin(nx)}{\sin x}$ is a regular function that can be written as a sum of cosines. $\endgroup$ Dec 16 '14 at 13:15
15
$\begingroup$

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: $$\int_{0}^{\pi}\log\sin x\,dx = \lim_{n\to +\infty}\frac{\pi}{n}\sum_{k=1}^{n-1}\log\sin\frac{\pi k}{n}=\lim_{n\to +\infty}\frac{\pi}{n}\log\frac{2n}{2^n}=\color{red}{-\pi\log 2}.$$

$\endgroup$
2
  • $\begingroup$ Hey, I lost it here $\cos((2n+1)x)=2\cos x\cos(2nx)-\cos((2n-1)x)$ would you care to explain how this is true? And how do you see that the integral is that series? $\endgroup$ Sep 8 '20 at 21:10
  • $\begingroup$ @PedroFernandes For the first question use the exponential representation of $\cos$: \begin{align*} \cos(n x) & = \frac{1}{2} \left( e^{i n x} - e^{- i n x}\right) \\ & = \frac{1}{2} \left( e^{i (n - 1) x} + e^{- i (n - 1) x}\right)\left(e^{i x} + e^{- i x}\right) - \frac{1}{2} \left( e^{i (n - 2) x} + e^{- i (n - 2) x}\right) \\ & = 2 \cos((n - 1) x) \cos(x) - \cos((n - 2) x). \end{align*} $\endgroup$
    – Ramanujan
    May 2 at 15:39
7
$\begingroup$

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 $\text{(2c)}$

$\endgroup$
3
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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.