I am looking for alternative solutions for finding this sum $$\sum_{n=1}^\infty \frac{\sin(nx)}{n} $$
My solution proceeds by integrating $$\sum_{n=1}^\infty e^{nx}=\frac{e^{ix}}{1-e^{ix}}$$ With suitable limits and then taking the imaginary part of it.
Note that\begin{align}\sum_{n=1}^\infty\frac{\sin(nx)}n&=\operatorname{Im}\left(\sum_{n=1}^\infty\frac{e^{inx}}n\right)\\&=\operatorname{Im}\left(\sum_{n=1}^\infty\frac{\left(e^{ix}\right)^n}n\right).\end{align}If $x\in\mathbb{R}\setminus\{0\}$, then $e^{ix}\in S^1\setminus\{1\}$, and therefore$$\sum_{n=1}^\infty\frac{\left(e^{ix}\right)^n}n=\log\left(\frac1{1-e^{ix}}\right),$$(where $\log$ is the principal branch of the logarithm), since\begin{align}\log(1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}x+\cdots&\implies\log(1-x)=-\left(x+\frac{x^2}2+\frac{x^3}3+\frac{x^4}4+\cdots\right)\\&\implies\log\left(\frac1{1-x}\right)=x+\frac{x^2}2+\frac{x^3}3+\frac{x^4}4+\cdots\end{align}Therefore\begin{align}\sum_{n=1}^\infty\frac{\sin(nx)}n&=-\arctan\left(\frac{-\sin x}{1-\cos(x)}\right)\\&=\arctan\left(\cot\left(\frac x2\right)\right).\end{align}Of course, if $x=0$, then this equality also holds.
