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It is known that one can sometimes derive certain Fourier series without alluding to the methods of Fourier analysis. It is often done using complex analysis. There is a way of deriving the formula $$\sum_{k = 1}^\infty \frac{\sin(kz)}{k} = \frac{\pi - z}{2}$$ using complex analysis for some $z$. In other words, it can be shown that $\sum_{k = 1}^\infty \sin(kz)/k$ converges to $(\pi - z)/2$ for some $z$. My question is: How can we show that the formula above holds for $z \in (0, 2\pi)$ without using Fourier analysis?

Edit:

Using blue's suggestion I concocted the following proof.

Let $\log z$ be the principal value of the logarithm (with the branch cut along the negative real axis). Recall that $e^{iz} - e^{-iz} = 2i\sin z$ and $\text{Arg}(z) = \text{Arg}(z^*)$. Furthermore, we have \begin{align*} \log(1 - e^{iz}) &= -\sum_{k = 1}^\infty \frac{e^{ikz}}{k} \\ &= \log|1 - e^{iz}| + i\theta \\ &= \log\left|2\sin\frac{z}{2}\right| + i\theta, \end{align*} where $\theta = \text{Arg}(1 - e^{iz})$. Now, write $1 - e^{iz} = 1 - \cos z - i\sin z$ and let $z \in (0, \pi)$. Then $$\tan \theta = \frac{\sin z}{1 - \cos z} = \frac{\cos(z/2)}{\sin(z/2)} = \tan\left(\frac{\pi}{2} - \frac{z}{2}\right).$$ Hence, $\theta = \pi/2 - z/2$. Moreover, since $z \in (0, \pi)$, we have $\theta \in (0, \pi/2)$. Using Euler's formula $e^{iz} = \cos z + i \sin z$ and the fact that the sine function is odd, we see that equating the imaginary parts gives $$\sum_{k = 1}^\infty \frac{\sin kz}{k} = \frac{\pi - z}{2}, \quad z \in (-\pi, 0) \cup (0, \pi).$$ Lastly, observe that $(-\pi, 0) \cup (0, \pi)$ can be replaced by $(0, 2\pi)$ due to the periodicity of the sine function.

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    $\begingroup$ This one is closely related to your question and it is easily evaluated by means of Abel-Plana Formula. $\endgroup$ – Felix Marin Jun 28 '14 at 2:03
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In order to make sense of sines with complex arguments, you need to decompose them into complex exponentials. After that notice you're working with two series, each a Taylor expansion of the natural logarithm function (written appropriately). Proceed from there.

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  • $\begingroup$ Thanks. I just added a proof using your suggestion. Is that what you meant? $\endgroup$ – glebovg Jun 28 '14 at 0:04
  • $\begingroup$ Is there a shorter way? $\endgroup$ – glebovg Jun 28 '14 at 0:52
  • $\begingroup$ @glebovg Yes that is what I meant. Dunno if it can be shortened, but it's the thing that immediately came to mind when you asked for non-Fourier arguments. $\endgroup$ – blue Jun 28 '14 at 2:53

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