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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?

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  • $\begingroup$ 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. $\endgroup$ – Dianbin Bao Feb 3 '14 at 22:38
  • $\begingroup$ That sounds very interesting. I'll check it out. $\endgroup$ – Peder Feb 3 '14 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. $\endgroup$ – Felix Marin Dec 15 '14 at 15:56
  • $\begingroup$ Possible duplicate of Compute the fourier coefficients, and series for $\log(\sin(x))$ $\endgroup$ – Alex M. Sep 3 '16 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 '16 at 20:17
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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 $\pmod {2 \pi i}$.

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  • $\begingroup$ Thank you. That was very elegantly done. $\endgroup$ – Peder Feb 2 '13 at 16:32
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    $\begingroup$ 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 '13 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 '13 at 18:07
  • $\begingroup$ @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 $|z|=|e^{i2x}|=1$. $\endgroup$ – Duchamp Gérard H. E. May 6 '18 at 17:05

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