# Fourier series of Log sine and Log cos

I saw the two identities $$-\log(\sin(x))=\sum_{k=1}^\infty\frac{\cos(2kx)}{k}+\log(2)$$ and $$-\log(\cos(x))=\sum_{k=1}^\infty(-1)^k\frac{\cos(2kx)}{k}+\log(2)$$ here: twist on classic log of sine and cosine integral. How can one prove these two identities?

• By the way, I saw on Kato's number theory book, there is an identity by Euler $\zeta(3)=\frac{2}{7}\pi^2\log 2+\frac{16}{7}\int_{0}^{\frac{\pi}{2}}x\log (\sin x)d x$. The above identity about $\log \sin x$ is used to prove Euler's identity. Feb 3, 2014 at 22:38
• That sounds very interesting. I'll check it out. Feb 3, 2014 at 23:49
• Initially, they are derived as identities when $\displaystyle x \in \left(\,0,{\pi \over 2}\,\right)$. Later on you can play with $\displaystyle\sin$ and/or $\displaystyle\cos$ properties to reuse them in another interval. Dec 15, 2014 at 15:56
• Possible duplicate of Compute the fourier coefficients, and series for $\log(\sin(x))$ Sep 3, 2016 at 10:06
• It's admirable that you found the duplication - however your reference pointer seems to be the duplicatee so to speak as my question was asked in 2013 while it was asked in 2014! Sep 7, 2016 at 20:17

Recall that $$\cos(2kx) = \dfrac{e^{i2kx} + e^{-i2kx}}2.$$ Hence, \begin{aligned}\sum_{k=1}^{\infty} \dfrac{\cos(2kx)}k &= \sum_{k=1}^{\infty} \dfrac{e^{i2kx} + e^{-i2kx}}{2k} \\&= \dfrac12 \big(-\log (1-e^{i2x} )-\log (1-e^{-i2x} ) \big) \\&= - \dfrac12 \log \big(2 - 2\cos(2x) \big) \\&= - \dfrac12 \log\big(4 \sin^2(x)\big) \\&= - \log 2 - \log\big(\sin(x)\big).\end{aligned} Hence, $$-\log\big(\sin(x)\big) = \sum_{k=1}^{\infty} \dfrac{\cos(2kx)}k + \log 2.$$ I leave it to you to similarly prove the other one. Both of these equalities should be interpreted $\pmod {2 \pi i}$.

• Thank you. That was very elegantly done. Feb 2, 2013 at 16:32
• I think you have two tiny mistakes in your evaluation that conveniently cancel out. Firstly, note that $\cos(2kx)=(e^{2ikx}+e^{-2ikx})/2$ and secondly note that $2-2\cos(x)=4\sin^2(x/2)$ Feb 2, 2013 at 16:48
• @Peder Ha ha! Thanks for pointing it out. Have corrected and updated it.
– user17762
Feb 2, 2013 at 18:07
• @user17762 Could you elaborate the "security" part of your (elegant, but formal) derivation (i.e. the domains) ? In particular, you implicitely use $$-\log(1-z)=\sum_{n\geq 1}\frac{z^n}{n}$$ on the border of the disk as $|z|=|e^{i2x}|=1$. May 6, 2018 at 17:05
• The Fourier series of $\ln( \cos x)$ can be obtained by setting $x\to \pi/2-x$ in the Fourier series of $\ln( \sin x)$ Jun 15, 2022 at 8:21

Here is another solution that addresses the concerns of Duchamp Gérard H. E.

We appeal to the following well known result in the theory of Fourier series:

Theorem: If $$f\in L_p(\mathbb{S}^1)$$, $$f\sim \sum_{n\in\mathbb{Z}}c_n e^{-in\theta}$$, and $$1\leq p<\infty$$, then the Abel sum $$A_rf=\sum _{n\in\mathbb{Z}}r^{|n|}c_ne^{in\theta}$$ converges to $$f$$ in $$L_p$$ and pointwise at every Lebesgue point of $$f$$ as $$r\nearrow1$$.

First from $$-\log(1-re^{i\theta})=\sum_{n\geq1}\frac{r^ne^{ni\theta}}{n}=-\log|1-re^{i\theta}| -i\operatorname{arg}(1-re^{i\theta})$$ where $$\log$$ is the principal branch of logarithm and $$0\leq r<1$$, we have that $$\sum_{n\geq1}\frac{r^n\cos n\theta}{n}=-\log|1-re^{i\theta}|\tag{1}\label{one}$$ $$\sum_{n\geq1}\frac{r^n\sin n\theta}{n}=-\operatorname{arg}\big(1-re^{i\theta}\big) \tag{2}\label{two}$$

The left-hand side of $$\eqref{one}$$ is the Abel sum of the series $$g(\theta)=\sum_{n\geq1}\frac{\cos n\theta}{n}=\frac{1}{2}\sum_{|n|\geq1}\frac{e^{in\theta}}{|n|}$$, a square integrable function.

It follows that $$\lim_{r\nearrow1}\sum_{n\geq1}\frac{r^n\cos n\theta}{n}=g(\theta)$$ at every Lebesgue point of $$g$$. On the other hand, $$\lim_{r\nearrow1}\log|1-re^{i\theta} |=|\log|1-e^{i\theta} |$$ for any $$0<\theta<2\pi$$. It follows that $$g(\theta)=\sum_{n\geq1}\frac{\cos n\theta}{n}=-\log|1-e^{i\theta} |$$ for all $$0<\theta<2\pi$$. As $$\log|1-e^{i\theta} |=\log\big(2\sin\frac{\theta}{2}\big)$$, we have that

$$\sum_{n\geq1}\frac{\cos n\theta}{n}=-\log 2 -\log\big(\sin\frac{\theta}{2}\big),\qquad 0<\theta< 2\pi\tag{3}\label{three}$$

Equation $$\eqref{two}$$ can be handle similarly. The left-hand side of being the Abel sum of the square integrable function $$h(\theta)=\sum_{n\geq1}\frac{\sin n\theta}{n}$$, converges to $$h(\theta)$$ at every Lebesgue point of $$h$$. It is well known that $$h(\theta)=\frac{1}{2}(\pi-\theta)$$ (the saw function) for $$0<\theta <2\pi$$. Hence

$$\sum_{n\geq1}\frac{\sin n\theta}{n}= -\operatorname{arg}(1-e^{i\theta})=\frac{1}{2}(\pi-\theta),\qquad 0<\theta< 2\pi\tag{4}\label{four}$$

In $$\eqref{three}$$, if $$0<\theta<\pi$$, then $$\pi<\theta<2\pi$$ and so,

\begin{aligned} -\log\Big(\cos\frac{\theta}{2}\Big)&=-\log\Big(\sin\big(\frac{\theta+\pi}{2}\big)\Big)\\ &=\log2 +\sum_{n\geq1}\frac{\cos(n(\theta+\pi))}{n}=\log2 +\sum_{n\geq1}\frac{(-1)^n\cos(n\theta)}{n} \end{aligned}