Compute the following integral over a closed curve Let $\gamma$ denote the unit circle center at origin. compute the following integral.
$$\int_\gamma \frac{e^z-e^{-z}}{z^4}dz$$
I guess I can solve this by integral by parts,
$$\int_\gamma \frac{e^z-e^{-z}}{z^4}dz=\int^{2\pi}_0 \frac{e^{e^{it}}-e^{-e^{it}}}{e^{3it}}dt=\int^{2\pi}_0 \frac{e^{e^{it}}}{e^{3it}}dt-\int^{2\pi}_0 \frac{e^{-e^{it}}}{e^{3it}}dt$$
we first compute $\int^{2\pi}_0 \frac{e^{e^{it}}}{e^{4it}}dt$. Let $u=e^{it}\implies du=ie^{it}dt$ and $dv=e^{3it}\implies v=e^{3it}/3i$,
hence we have $$[\frac{e^{it}e^{3it}}{3i}]^{2\pi}_0-\frac{1}{3}\int^{2\pi}_0e^{4it}dt=0$$
Similarily, $\int^{2\pi}_0 \frac{e^{-e^{it}}}{e^{3it}}dt=0$ Hence, we have the integral to be zero.
But I was wondering if I can use any propositions relating to holomorphic functions on a disc to conclude that the above integral is zero since $\gamma $ is a closed curve. If I want to compute the integral this way, how should I begin?
Thanks!

Using the residue theorem,  we get $2\pi i/3$, because the residue is $1/3$, the coefficient of the $z^3$ term of $e^z-e^{-z}$.
(I tried to correct your integration by parts, without success.)
