0

$$\log(\sin x)=-\log 2-\sum_{k\geq 1}\frac{\cos(2kx)}{k} \phantom{a} (0<x<\pi)$$

Consider:

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

And lets focus on the RHS first.

Using Abel-Plana Formula

$$\sum_{n=0}^{\infty} \frac{\cos(2nx)}{n} = \int_{0}^{\infty}\frac{\cos(2kx)}{k} dk + \frac{1}{2} f(0) + i\int_{0}^{\infty} \frac{f(it) - f(-it)}{e^{2\pi t} - 1} dt $$

But that doesnt converge?

Lebes
  • 1,636

1 Answers1

1

Hint: Use Euler's formula $e^{ix}=\cos x+i\sin x$ and the series expansion for $\ln(1+x)$.

Edit: Let $$S=\sum_{k\ge 1}\frac{\cos 2kx}{k}=\Re(T)\\ T=\sum_{k\ge 1}\frac{e^{-i2kx}}{k}=-\ln(1-e^{-i2x})=-\ln (2i\sin xe^{-ix})\\=-\ln \sin x-\ln 2-\ln(ie^{-ix})$$Now, $$\ln(ie^{-ix})=\ln(i\cos x+\sin x)=\ln 1+i(\pi/2-x+2n\pi),\ n\in \mathbb{Z}\\ \implies S=\Re(T)=-\ln \sin x-\ln 2$$

Samrat Mukhopadhyay
  • 16,277
  • Thanks (+1). But I do not understand something. I got:

    $$\sum_{n=1}^{\infty} (-1)^{n-1}x^{n-1} = \frac{1}{1+x}$$ Then:

    $$\sum_{n=1}^{\infty} (-1)^{n-1}\frac{x^n}{n} = \log(1+x)$$

    Then what should I do?

    (1) Moreover why $\log(1+x)$ instead of $\log(X)$? Thanks

     – Lebes Feb 22 '15 at 09:16
  • Or, $$\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$$

    What should I do?

     – Lebes Feb 22 '15 at 09:29
  • Ok, let me give you a proof to explain the steps. – Samrat Mukhopadhyay Feb 22 '15 at 11:50