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 $y$. $\endgroup$ – Mark May 4 '16 at 6:53
  • $\begingroup$ @Mark Can you please elaborate and post as answer? $\endgroup$ – Henry May 4 '16 at 6:55
  • $\begingroup$ Do we know something about $y$? $\endgroup$ – Hetebrij May 4 '16 at 7:57
  • $\begingroup$ @Hetebrij $x$ and $y$ are independent. $\endgroup$ – Henry May 4 '16 at 7:58
  • $\begingroup$ But do you want all $y \neq 0$, $y$ postive, $y$ bigger than $1$? $\endgroup$ – Hetebrij May 4 '16 at 8:02

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.