Show that : $ \sum^{\infty}_{n=1} \frac{ \sin(n \alpha)}{n} = \frac{\pi - \alpha}{2} $ Show that : 
$ \sum^{\infty}_{n=1} \frac{ \sin(n \alpha)}{n} = \frac{\pi - \alpha}{2} $
(Preferably) using complex analysis tools.
Any hints or ideas is appreciated.
 A: Hint : compute first $\sum_{k=1}^N \frac{\sin(k\alpha)}{n}$ by differentiating and noticing that one can recognize the real part of a geometric series.
A: Say we just want to sum that series, use,
$$ \sin(n\alpha) = \frac{e^{in\alpha} - e^{-in\alpha}}{2i}$$
Then use logarithmic series, 
$$\frac1{2i} \sum_{n=1}^{\infty} \frac{e^{ni\alpha}}{n} - \frac{e^{-ni\alpha}}{n} $$
Thus, we get, 
$$ \frac{1}{2i} \left\{ -\log(1-e^{i\alpha}) + \log(1-e^{-i\alpha}) \right\} $$
Now you got to simply that using identities and log rules. 
A: The sum can be evaluated by contour integration by considering $$ f(z) = \frac{\pi e^{i \alpha z} e^{- i \pi z} \csc (\pi z)}{z}$$ and integrating around a square with vertices at $\pm (N+1/2) \pm (N+1/2)$ where $N$ is a positive integer (call it $C_{N})$.
Like $\pi \cot \pi z$, $\pi e^{- i \pi z} \csc(\pi z)$ has simple poles at the integers with residue $1$.
But unlike $ \displaystyle \pi \int_{C_{N}} \frac{e^{i \alpha z}  \cot(\pi z)}{z} \ dz $, the integral $\displaystyle \pi \int_{C_{N}} \frac{e^{i \alpha z} e^{- i \pi z} \csc(\pi z)}{z}\ dz$ vanishes as $N \to \infty$ if $ 0 < \alpha < 2 \pi$.
Then
$$ \lim_{N \to \infty}\pi \int_{C_{N}} \frac{e^{i \alpha z} e^{- i \pi z} \csc(\pi z)}{z}\ dz = 0 = \sum_{n=-\infty}^{1} \text{Res}[f(z),n] + \text{Res}[f(z),0] + \sum_{n=1}^{\infty} \text{Res}[f(z),n]$$
$$ = \sum_{n=-\infty}^{-1} \frac{e^{i \alpha n}}{n} + \text{Res}[f(z),0] + \sum_{n=1}^{\infty} \frac{e^{i \alpha n}}{n}$$
And since
$$ f(z) = \frac{\pi e^{i(\alpha - \pi)z} \csc (\pi z)}{z} = \pi \Big( 1 + i(\alpha - \pi)z + \mathcal{O}(z^{2}) \Big) \Big(\frac{1}{\pi z} + \mathcal{O}(1) \Big)\frac{1}{z}$$
$$ = \frac{1}{z^{2}} + i \frac{\alpha - \pi}{z} + \mathcal{O}(1) $$
we have
$$\text{Res}[f(z),0] = i(\alpha- \pi) $$
Therefore,
$$ \sum_{n=-\infty}^{-1} \frac{e^{i \alpha n}}{n} + \sum_{n=1}^{\infty} \frac{e^{i \alpha n}}{n} = i(\pi -\alpha)  $$
And equating the imaginary parts on both sides of the equation,
$$ 2 \sum_{n=1}^{\infty} \frac{\sin (\alpha n)}{n} = \pi -\alpha $$
$$ \implies \sum_{n=1}^{\infty} \frac{\sin (\alpha n)}{n} = \frac{\pi - \alpha}{2}$$
