calculus - Evaluate $sum_{r=1}^{infty} frac{sin(rpi x)}{r cdot y^r}$

Find a closed form expression for

$$\sum_{r=1}^{\infty} \dfrac{\sin(r\pi x)}{r \cdot y^r}$$

I know that $\displaystyle\sum_{r=1}^{\infty} \dfrac{\sin(r \pi x)}{r} = \dfrac{\pi}{2} - \left\{\dfrac{x}{2}\right \}$ but I don't know how to obtain a closed form for the required summation. I thought about using Euler's Formula but it became messy.

Any help will be appreciated.
Thanks.

Answer

$$
\sum_{r=1}^{\infty} \dfrac{\sin(r\pi x)}{r \cdot y^r}=\mathrm{Im}\sum_{r=1}^{\infty} \dfrac{\exp(\mathrm{i}r\pi x)}{r \cdot y^r}=\mathrm{Im}\int_0^\infty ds \sum_{r=1}^{\infty} \left(\frac{e^{\mathrm{i}\pi x-s}}{y}\right)^r=\mathrm{Im}\int_0^\infty ds\frac{e^{i \pi x}}{e^s y-e^{i \pi x}}=
$$
$$
=-\mathrm{Im}\log \left(1-\frac{e^{i \pi x}}{y}\right)=- \mathrm{arctan}\left(1-\frac{\cos (\pi x)}{y},-\frac{\sin (\pi x)}{y}\right)\ ,
$$
where $\log$ is the principal branch of the complex logarithm, and we used $1/z=\int_0^\infty ds\ e^{-s z}$, for $z>0$. The function arctan with two arguments is described here https://reference.wolfram.com/language/ref/ArcTan.html. I checked with Mathematica a few cases and it seems it works.