Find all possible values for $$\int_{\Gamma}\frac{e^{z}dz}{z(z^{2}-1)}$$ as $\Gamma$ is a rectangle contour that does not passes through the points $0,1,-1$.
My attempt:
we must divide the answer into cases.
Notice that given $\Gamma$ to be a rectangle, it can't be that $-1,1\in \Gamma \land 0\notin \Gamma$
Case 1: $-1,0,0 \notin \Gamma$
By Cauchy theorem, $$\int_{\Gamma}\frac{e^{z}dz}{z(z^{2}-1)}= 0$$
for all the following cases we have to consider the direction of the contour, right or left, which means the sign (positive or negative) of the integrated.
Case $2\space\text{and}\space 3$: $-1\in \Gamma^{0}, 0,1\notin \Gamma$
Let $\displaystyle f(z)=\frac{e^z}{z(z-1)}$, then $$\int_{\Gamma}\frac{e^{z}dz}{z(z^{2}-1)} = \int_{\Gamma}\frac{f(z)dz}{z+1} = \pm 2\pi i(f(-1))=\pm\frac{\pi i}{e}$$
Case $4\space\text{and}\space 5$: $0\in \Gamma, 1,-1\notin \Gamma$
Let $\displaystyle f(z)=\frac{e^z}{z^2-1}$, then $$\int_{\Gamma}\frac{e^{z}dz}{z(z^{2}-1)} = \int_{\Gamma}\frac{f(z)dz}{z} = \pm 2\pi i(f(0))=\pm(2\pi i)$$
Case $ 6 \space\text{and}\space 7$: $1\in \Gamma^{0}, 0,-1\notin \Gamma$
Let $\displaystyle f(z)=\frac{e^z}{z(z+1)}$, then $$\int_{\Gamma}\frac{e^{z}dz}{z(z^{2}-1)} = \int_{\Gamma}\frac{f(z)dz}{z(z+1)} = \pm 2\pi i(f(1))=(\frac{e^{1}}{1(2)}) = \pm{\pi i e}$$
Case $8\space\text{and}\space 9$: $-1,0\in \Gamma, 1\notin \Gamma$
Let $\displaystyle f(z)=\frac{e^z}{(z-1)}$, then $$\int_{\Gamma}\frac{e^{z}dz}{z(z^{2}-1)} = \int_{\Gamma}\frac{f(z)dz}{z(z+1)} = \pm 2 \pi i(f(-1)+f(0))= \pm 2 \pi i (\frac{e^{-1}}{-2} + 1) = \pm (\frac{-\pi i}{e}+ 2\pi i)$$
Case $10\space\text{and}\space 11$: $1,0\in \Gamma, -1\notin \Gamma$
Let $\displaystyle f(z)=\frac{e^z}{(z+1)}$, then $$\int_{\Gamma}\frac{e^{z}dz}{z(z^{2}-1)} = \int_{\Gamma}\frac{f(z)dz}{z-1} = \pm 2 \pi i(f(1)+f(0))= \pm 2 \pi i (\frac{e^{1}}{2} + 1) = \pm (\pi i e+ 2\pi i)$$
Case $12\space\text{and}\space13$: $1,0,-1, \in \Gamma$
Let $\displaystyle f(z)={e^z}$, then $$\int_{\Gamma}\frac{e^{z}dz}{z(z^{2}-1)} = \int_{\Gamma}\frac{f(z)dz}{z(z^2-1)} = \pm 2 \pi i(f(1)+f(0)+f(-1))= \pm 2 \pi i (e^{1} + e^{-1} + 1)$$
is that correct?