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!

You can directly use Cauchy's differentiation formula:
$$f^{(n)}(a) = \frac{n!}{2\pi i}\oint_{\gamma}\frac{f(z)}{(z-a)^{n+1}}$$
where
$$a=0, f(z) = e^z-e^{-z} = 2\sinh z \text{ and } \gamma: |z|=1$$
Just noting that $\sinh^{(3)} z = \cosh z$ you get
$$\int_\gamma \frac{e^z-e^{-z}}{z^4}dz = \frac{2\pi i}{3!}\cdot 2\cosh 0 = \frac 23 \pi i$$
