How to compute the complex integral $\int_{\gamma} e^{\frac{1}{z^2 - 1}}\sin{\pi z} \, \mathrm dz$?

Using your series expansions we can write: $$ e^{\frac1{z^2-1}} \sin \pi z = \sum_{n,k \ge 0} (-1)^{k+1} \frac{\pi^{2k+1}}{(2k+1)!\,n!} \frac1{(z+1)^n}\frac1{(z-1)^{n-2k-1}} $$ Now (for any $z$ inside the circle of radius 2 centered at $z=1$) $$ (z+1)^{-n} = \sum_{l\ge0} \frac{(-n)(-n-1)\cdot\ldots\cdot(-n-l+1)}{l!}2^{-n-l} (z-1)^l $$ So in total we get $$ e^{\frac1{z^2-1}} \sin \pi z = \sum_{n,k,l \ge 0} (-1)^{k+1} \frac{\pi^{2k+1}}{(2k+1)!\,n!} {-n \choose l} \frac1{2^{n+l}} \frac1{(z-1)^{n-2k-l-1}} $$ The coefficient of $\frac1{z-1}$ is then $$ \sum_{n,k \ge 0} \frac{(-1)^{k+1} \pi^{2k+1}}{(2k+1)!\,n!\,4^{n-k}} {-n \choose n-2k} $$ There must be a nicer solution however....