Difficult infinite trigonometric series

Solution 1:

$\displaystyle\sum_{n=1}^\infty\frac{\sin(nx)}n=$Img$\left(\sum_{n=1}^\infty\dfrac {(e^{ix})^n}n\right)$

Now, $\displaystyle\sum_{n=1}^\infty\dfrac {(e^{ix})^n}n=-\ln(1-e^{ix})$

Again, $\displaystyle1-e^{2iy}=-e^{iy}(e^{iy}-e^{-iy})=-e^{iy}2i\sin y$

$\displaystyle\implies\ln(1-e^{2iy})=\log2+iy+\ln(\sin y)+\text{Log}(-i)$

and $\displaystyle\text{Log}(-i)=\text{Log}\left[\cos\left(-\frac\pi2\right)+i\sin\left(-\frac\pi2\right)\right]=\text{Log}_ee^{\left(2n\pi -\frac\pi2\right)i}$

$\displaystyle\implies\text{Log}(-i)=\left(2n\pi -\frac\pi2\right)i$

Here $\displaystyle 2y=x=\frac\pi3$

Solution 2:

The result is none other than $\dfrac\pi3$ , with the obvious generalization $\displaystyle\sum_{n=1}^\infty\frac{\sin(nx)}n=\frac{\pi-x^*}2$, where $x^*=x\mod2\pi.~$ Unfortunately, the only way in which I am able to “prove” this is by ascribing finite values to divergent series by way of analytic continuation, which isn't really “kosher”, so $...$