Evaluate $\sum\limits_{n\mathop=1}^{\infty}\frac{\sin2\pi nx}{\pi n}$

Prove $$x-\lfloor x\rfloor-\frac{1}{2}=-\sum_{n\mathop =1}^{\infty}\frac{\sin(2\pi nx)}{\pi n}$$ where $x$ is any non-integer real number.

• Every term of the series has $1$ as a period. Therefore the sum of the series (if it exists, which it does) must have $1$ as a period. For proving convergence, distinguish between $x\in\mathbb{Z}$ and $x\notin\mathbb{Z}$, and for the latter, use the Dirichlet test. – Daniel Fischer Apr 15 '14 at 23:37
• Why do you think your answer is wrong? – Mhenni Benghorbal Apr 15 '14 at 23:41
• Two related computations (though somewhat more complicated) are at these two links: MSE link I and MSE link II. – Marko Riedel Apr 15 '14 at 23:53

$$\sin t=\Im(e^{it})\quad=>\qquad\frac{\sin(2\pi nx)}{2\pi n}=\Im\bigg(\frac{e^{2\pi inx}}{2\pi n}\bigg)=\Im\int e^{2\pi inx}dx\qquad=>$$
$$\sum_1^\infty\frac{\sin(2\pi nx)}{2\pi n}=\Im\int\sum_1^\infty e^{2\pi inx}dx=\Im\int\frac{e^{2\pi ix}}{1-e^{2\pi ix}}dx=-\Im\bigg[\frac{\ln(1-e^{2\pi ix})}{2\pi}\bigg]$$