TITLE: Calculate infinite summation of sin(x)/x
QUESTION [5 upvotes]: How do you calculate the infinite sum $\sum_{i=1}^{\infty} \frac{\sin(i)}{i}$? According to Wolfram Alpha, the value of the sum is $\frac{\pi - 1}{2}$, but it does not tell me the method by which it gets this result.

REPLY [2 votes]: Copied from this answer:

Using the power series
  $$
-\log(1-z)=\sum_{k=1}^\infty\frac{z^k}{k}
$$
  we get
  $$
\begin{align}
\sum_{k=1}^\infty\frac{\sin(k)}{k}
&=\frac1{2i}\sum_{k=1}^\infty\frac{e^{ik}-e^{-ik}}{k}\\
&=\frac1{2i}\left[-\log(1-e^i)+\log(1-e^{-i})\right]\\
&=\frac1{2i}\log(-e^{-i})\\
&=\frac{\pi-1}{2}
\end{align}
$$
  That is, since $1-e^{-i}$ is in the first quadrant and $1-e^i$ is in the fourth, the imaginary part of $-\log(1-e^i)+\log(1-e^{-i})$ is between $0$ and $\pi$.

Convergence is guaranteed by Dirichlet's Test and convergence to the value expected by Abel's Theorem.