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.

  • $\begingroup$ It may be useful to use a summation by parts, between the sum you already know, and the "other part", which is a geometric series in $y$. $\endgroup$ May 4, 2016 at 6:53
  • $\begingroup$ @Mark Can you please elaborate and post as answer? $\endgroup$
    – Henry
    May 4, 2016 at 6:55
  • $\begingroup$ Do we know something about $y$? $\endgroup$
    – Hetebrij
    May 4, 2016 at 7:57
  • $\begingroup$ @Hetebrij $x$ and $y$ are independent. $\endgroup$
    – Henry
    May 4, 2016 at 7:58
  • $\begingroup$ But do you want all $y \neq 0$, $y$ postive, $y$ bigger than $1$? $\endgroup$
    – Hetebrij
    May 4, 2016 at 8:02

1 Answer 1


$$ \sum_{r=1}^{\infty} \dfrac{\sin(r\pi x)}{r \cdot y^r}=\mathrm{Im}\sum_{r=1}^{\infty} \dfrac{\exp(\mathrm{i}r\pi x)}{r \cdot y^r}=\mathrm{Im}\int_0^\infty ds \sum_{r=1}^{\infty} \left(\frac{e^{\mathrm{i}\pi x-s}}{y}\right)^r=\mathrm{Im}\int_0^\infty ds\frac{e^{i \pi x}}{e^s y-e^{i \pi x}}= $$ $$ =-\mathrm{Im}\log \left(1-\frac{e^{i \pi x}}{y}\right)=- \mathrm{arctan}\left(1-\frac{\cos (\pi x)}{y},-\frac{\sin (\pi x)}{y}\right)\ , $$ 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.


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