How to manage an essential singularity?
$$I=\oint_{\gamma} e^{\frac{1}{z}}z^{-3}dz$$
where gamma= circumference centered in z=0, radius=2.
Clearly z=0 is an essential singularity, so I tried to use the Laurent series..
$$I=\oint \frac{1}{z^3}\sum_0 ^\infty (\frac{1}{z})^n \frac{1}{n!}=\oint \sum_0 ^\infty \frac{1}{z^{n-3}} \frac{1}{n!}$$
I tried to obtain the $a_{-1}$ coefficient of the series.. that corresponds to the residual... so the value of the integral should be $$I=2\pi i a_{-1} $$.. but I'm stuck, because if n=-1, I get the factorial of a negative number...
I get $$z^{-3}e^{1/z}=\frac1{z^3}+\frac1{z^4}+\frac1{2z^5}+\cdots.$$ The residue at $0$ (the $z^{-1}$ coefficient) is zero, so the contour integral is zero.
Integrating $z^3e^{1/z}$ instead gives a more interesting answer!