# Evaluate the integral $\int\exp(\frac{z^2-3}{z^3})$

evaluate the integral $\int\exp(\frac{z^2-3}{z^3})$ the unit circle oriented counterclockwise.

I think that we use the following theorem:

Suppose that $f$ is holomorphic in an open set containing a circle $C$ and its interiror, except for poles at the points $z_1,...,z_N$ inside $C$. Then $$\int_{C}f(z)dz=2\pi i\sum_{k=1}^{N}Res(f,z_k)$$

we have only pole $z_0=0$ for $f(z)=\exp(\frac{z^2-3}{z^3})$ and

$$Res(f,0)=\frac{1}{2}\frac{d^2}{dz^2}\lim_{z\to0}z^3f(z)$$

Is there another way to evaluate this integral? Thanks.

## 1 Answer

The residue method you suggest is fine. But you may not calculate it by the stated formula. The problem is that there is an essential singularity at the origin.

Hint: Several ways to go, but you may e.g. write: $$\exp (\frac{z^2-3}{z^3}) = \exp(\frac{1}{z}) \exp(\frac{-3}{z^3})$$ Do a Laurent series expansion (if you know about those?) and pick up the coefficient to $1/z$. [it is in fact quite simple]