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?

  • $\begingroup$ 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. $\endgroup$ Feb 3, 2014 at 22:38
  • $\begingroup$ That sounds very interesting. I'll check it out. $\endgroup$
    – Peder
    Feb 3, 2014 at 23:49
  • $\begingroup$ 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. $\endgroup$ Dec 15, 2014 at 15:56
  • $\begingroup$ Possible duplicate of Compute the fourier coefficients, and series for $\log(\sin(x))$ $\endgroup$
    – Alex M.
    Sep 3, 2016 at 10:06
  • $\begingroup$ 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! $\endgroup$
    – Peder
    Sep 7, 2016 at 20:17

2 Answers 2


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 $\pmod {2 \pi i}$.

  • $\begingroup$ Thank you. That was very elegantly done. $\endgroup$
    – Peder
    Feb 2, 2013 at 16:32
  • 2
    $\begingroup$ 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 $2-2\cos(x)=4\sin^2(x/2)$ $\endgroup$
    – Peder
    Feb 2, 2013 at 16:48
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    $\begingroup$ @Peder Ha ha! Thanks for pointing it out. Have corrected and updated it. $\endgroup$
    – user17762
    Feb 2, 2013 at 18:07
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    $\begingroup$ @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 $|z|=|e^{i2x}|=1$. $\endgroup$ May 6, 2018 at 17:05
  • $\begingroup$ The Fourier series of $\ln( \cos x)$ can be obtained by setting $x\to \pi/2-x$ in the Fourier series of $\ln( \sin x)$ $\endgroup$ 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}


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