Find the residue of $e^{\frac{1}{z^2-1}}\sin(\pi z)$ at $z=1$ I'm dealing with the following problem (from an old qualifying exam):

Let $\gamma$ be a closed curve in the right half-plane that has index $N$ with respect to the point 1. Find
  $$
\int_{\gamma}e^{\frac{1}{z^2-1}}\sin(\pi z)\,dz.
$$

I take this problem as, "Find $\text{Res}\left(f,1\right)$," where $f$ is the integrand. For an essential singularity like I believe $f$ has at $z=1$, you have to expand into the Laurent series. This is the closest I got: By partial fraction decomposition,
$$
\frac{1}{z^2-1}=\frac{1}{2(z-1)}-\frac{1}{2(z+1)},
$$
So
$$
f(z)=e^{\frac{1}{2(z-1)}}e^{-\frac{1}{2(z+1)}}\sin(\pi z)=\frac{1}{2i}e^{\frac{1}{2(z-1)}}\sin\left(\frac{z\pi}{2(z+1)}\right).
$$
If I could magically find a Taylor expansion about $z=1$ of $\sin\left(\frac{z\pi}{2(z+1)}\right)$ I'd be all set, but it seems quite disgusting to work out. I've also tried integrating over a simpler closed curve containing $z=1$ but I can't seem to find something that turns out integrable. Any help is greatly appreciated. Thanks.
 A: Expanding the functions $\exp$ and $\sin$ in Laurent series we get
[
\begin{align}
 \exp \left( {\frac{1}{{z^2  - 1}}} \right) = \sum {\frac{1}{{\left( {z^2  - 1} \right)^n n!}}}  = \sum {\frac{1}{{\left( {z + 1} \right)^n \left( {z - 1} \right)^n n!}}}  = \sum {\frac{{\frac{1}{{\left( {z + 1} \right)^n }}}}{{\left( {z - 1} \right)^n n!}}}  
\end{align}
Also,
\begin{align}
 \sin \pi z = \sin \pi \left( {z + 1 - 1} \right)=\sin \left( {\pi \left( {z - 1} \right) + \pi } \right) = \sin \pi \left( {z - 1} \right)\cos \pi  + \sin \pi \cos \pi \left( {z - 1} \right) \\ 
  =  - \sin \pi \left( {z - 1} \right) 
 \end{align}
and since 
\begin{align}
\sin z = \sum {\frac{{\left( { - 1} \right)^n }}{{\left( {2n + 1} \right)!}}z^{2n + 1} }  \Rightarrow \sin \pi \left( {z - 1} \right) = \sum {\frac{{\left( { - 1} \right)^n }}{{\left( {2n + 1} \right)!}}\pi ^{2n + 1} \left( {z - 1} \right)^{2n + 1} } 
\end{align}
Thus,
\begin{align}
&\exp \left( {\frac{1}{{z^2  - 1}}} \right)\sin \pi \left( {z - 1} \right)
\\ 
&=  - \left( {\sum {\frac{1}{{\left( {z + 1} \right)^n \left( {z - 1} \right)^n n!}}} } \right)\left( {\sum {\frac{{\left( { - 1} \right)^n }}{{\left( {2n + 1} \right)!}}\pi ^{2n + 1} \left( {z - 1} \right)^{2n + 1} } } \right) \
\\
&=- \left( {1 + \frac{{\frac{1}{{\left( {z + 1} \right)}}}}{{\left( {z - 1} \right)}} + \frac{{\frac{1}{{\left( {z + 1} \right)^2 }}}}{{\left( {z - 1} \right)^2 2!}} + \frac{{\frac{1}{{\left( {z + 1} \right)^3 }}}}{{\left( {z - 1} \right)^3 3!}} + \frac{{\frac{1}{{\left( {z + 1} \right)^4 }}}}{{\left( {z - 1} \right)^4 4!}} + \frac{{\frac{1}{{\left( {z + 1} \right)^5 }}}}{{\left( {z - 1} \right)^5 5!}}  +   \cdots } \right) \\ 
  &\times \left( {\pi \left( {z - 1} \right) - \frac{{\pi ^3 }}{{3!}}\left( {z - 1} \right)^3  + \frac{{\pi ^5 }}{{5!}}\left( {z - 1} \right)^5  +  \cdots   +  \cdots } \right) \\ 
 &\Rightarrow \text{The coefficient of }\frac{1}{z-1}=- 
\left( {\frac{\pi }{{1!2!}}\frac{1}{{\left( {z + 1} \right)^2 }} - \frac{{\pi ^3 }}{{3!4!}}\frac{1}{{\left( {z + 1} \right)^4 }} + \frac{{\pi ^5 }}{{5!6!}}\frac{1}{{\left( {z + 1} \right)^6 }} +  \ldots  } \right)
\\
&=
 - \sum {\frac{{\left( { - 1} \right)^n \pi ^{2n + 1} }}{{n!\left( {n + 1} \right)!}}\frac{1}{{\left( {z + 1} \right)^{2n + 2} }}} 
  = g\left( z \right)
\end{align}
\begin{align}
{\mathop{\rm Res}\nolimits} \left\{ {f\left( z \right),1} \right\} = g\left( 1 \right)
 = 
 - \sum {\frac{{\left( { - 1} \right)^n \pi ^{2n + 1} }}{{n!\left( {n + 1} \right)!}}\frac{1}{{2^{2n + 2} }}}  
\end{align}
