Calculate infinite summation of sin(x)/x 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. 
 A: 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.
A: Consider the two summations $$S=\sum_{k=1}^{\infty} \frac{\sin(k)}{k}\qquad \text{and} \qquad C=\sum_{k=1}^{\infty} \frac{\cos(k)}{k}$$
$$C+iS=\sum_{k=1}^{\infty} \frac{\cos(k)+i \sin(k)}{k}=\sum_{k=1}^{\infty} \frac{e^{i k}} k=\sum_{k=1}^{\infty} \frac{(e^{i})^ k} k=-\log \left(1-e^i\right)$$
$$C-iS=\sum_{k=1}^{\infty} \frac{\cos(k)-i \sin(k)}{k}=\sum_{k=1}^{\infty} \frac{e^{-i k}} k=\sum_{k=1}^{\infty} \frac{(e^{-i})^ k} k=-\log \left(1-e^{-i}\right)$$ Expanding the logarithm
$$C+i S=-\frac{1}{2} \log \left(\sin ^2(1)+(1-\cos (1))^2\right)+i \tan
   ^{-1}\left(\frac{\sin (1)}{1-\cos (1)}\right)$$ $$C+i S=-\frac{1}{2} \log (2-2 \cos (1))+i \left(\frac{\pi }{2}-\frac{1}{2}\right)$$
$$C-i S=-\frac{1}{2} \log \left(\sin ^2(1)+(1-\cos (1))^2\right)-i \tan
   ^{-1}\left(\frac{\sin (1)}{1-\cos (1)}\right)$$
$$C-i S=-\frac{1}{2} \log (2-2 \cos (1))+i \left(\frac{1}{2}-\frac{\pi }{2}\right)$$
$$C=\frac{(C+iS)+(C-iS)}2=-\frac{1}{2} \log (2-2 \cos (1))$$
$$S=\frac{(C+iS)-(C-iS)}{2i}=\frac{1}{2} (\pi -1)$$
