Evaluating - $\int_{|z| =R,R > 0} z^2 \big( \sin\big(\frac{1}{z}\big) + \sin\big(\frac{2}{z}\big) + ...+ \sin\big(\frac{n}{z}\big) \big) dz $ Let:
$$ I_n =\int_{|z| =R , \text{ }R > 0} z^2 \left( \sin\left(\frac{1}{z}\right) + \sin\left(\frac{2}{z}\right) + ...+ \sin\left(\frac{n}{z}\right) \right)\, dz  $$
Knowing that $n$ is a natural non-zero number, prove that $e^{12I_n}$ is an integer. This was a problem on local contest today and I have no idea how to solve this.
Can anyone help?

\begin{eqnarray}
I_n&=&\sum_{k=1}^n\int_{|z|=R}z^2\sin\left(\frac{k}{z}\right)\,dz=\sum_{k=1}^n\sum_{m=0}^\infty\int_{|z|=R}z^2\frac{(-1)^m}{(2m+1)!}\frac{k^{2m+1}}{z^{2m+1}}\,dz\\
&=&\sum_{k=1}^n\sum_{m=0}^\infty\int_{|z|=R}\frac{(-1)^m}{(2m+1)!}\frac{k^{2m+1}}{z^{2m-1}}\,dz=\sum_{k=1}^n\int_{|z|=R}\frac{-k^3}{3!}\frac{1}{z}\,dz=-\frac13i\pi\sum_{k=1}^nk^3\\
&=&-\frac13i\pi\left(\sum_{k=1}^nk\right)^2=-\frac13i\pi\left[\frac{n(n+1)}{2}\right]^2=-i\frac{n^2(n+1)^2}{12}
\end{eqnarray}
