# Laurent series for $\exp(-1/z)\sin(1/z)$ at $z=0$

I am trying to evaluate the integral $$\int \exp\left(-\frac{1}{z}\right)\sin\left(\frac{1}{z}\right)dz$$ in the deleted neighborhood $|z|=1$. This integral can be easily solved using the Cauchy integral formula, but this requires figuring out the the Laurent series since $z=0$ is an essential singularity. Could someone please show me how to write the Laurent series in order to find the $a_{-1}$ term. Thank you

• writing $\sin (1/z) = (e^{i/z} - e^{-i/z})/(2i)$ and using expression of the exponential function. Apr 17, 2017 at 21:24

$\exp(-w) \sin(w)$ is an entire function. It has Maclaurin series $$\exp(-w)\,\sin(w) = w - w^2 + \frac{1}{3}\,w^3 + \dots$$ valid in the whole complex plane. Therefore your function has a convergent Laurent series $$\exp\left(-\frac{1}{z}\right)\sin\left(\frac{1}{z}\right) = \frac{1}{z} - \frac{1}{z^2} +\frac{1}{3z^3} +\dots$$ valid for all $z \ne 0$.
• Thank you for the response, now I understand. $a_{-1}$=1 . Apr 18, 2017 at 12:20