TITLE: Evaluate $\sum_{n=1}^\infty\frac{\sin(n)}{n^3}$
QUESTION [6 upvotes]: I want to find $\displaystyle\sum_{n=1}^\infty\frac{\sin(n)}{n^3}$ by using method from this link :
http://www.supermath.info/InfiniteSeriesandtheResidueTheorem.pdf
Since $f(z)=\frac{\sin{z}}{z^3}$ has a pole at $z=0$,
I get that $\displaystyle\sum_{n=-\infty ,n\neq0}^\infty\frac{\sin(n)}{n^3}=-Res_{z=0}\left(\frac{\pi\sin{z}\cot{(\pi z)}}{z^3}\right)=\frac{2\pi^2+1}{6}$
.
That is $\displaystyle\sum_{n=1}^\infty\frac{\sin(n)}{n^3}=\frac{2\pi^2+1}{12}$.
But wolfram compute that $\displaystyle\sum_{n=1}^\infty\frac{\sin(n)}{n^3}=\frac{(\pi-1)(2\pi-1)}{12}$.
https://www.wolframalpha.com/input/?i=sum+sin(n)%2Fn%5E3,+n%3D1+to+inf
How am I wrong? 
Thank you.

REPLY [3 votes]: You can not apply the approach from that paper directly as the key assumption of theorem 3.2 
$$
|f(z)|\le \frac{M}{|z|^k}
$$
for some $k>1$ and all large $z$, $|z|>R$ for some large $R$, is not satisfied for $f(z)=\frac{\sin z}{z^3}$ as the sine function is exponential in the imaginary directions,
$$
\sin(z)=\sin(x+iy)=\sin(x)\cosh(y)+i\cos(x)\sinh(y)
$$