I'm new with these complex contour integrals and I wanted to know if my approach to this one is correct. The integral is $$\oint_\gamma \frac{\sin z}{z^3-z}\,\mathrm{d}z$$ where $\gamma(t)=2e^{it}, t\in[0,2\pi]$. First, the integrand $f(z)$ has three singularities in the region enclosed by $\gamma$:
- $z=0$ that results a removable singularity
- $z=1$, pole of order $1$
- $z=-1$, pole of order $1$
Since the Residue Theorem involves only poles, I computed $\mathrm{Res}f|_{-1}=-\frac{\sin 1}2$ and $\mathrm{Res}f|_1=\frac{\sin 1}2$. Hence $$\oint_\gamma \frac{\sin z}{z^3-z}\,\mathrm{d}z=2\pi i(\mathrm{Res}f|_{-1}+\mathrm{Res}f|_1)=0$$ I think my calculation are right but I have a doubt about the Laurent series of $f(z)$ centered in each singularity I found… Actually, I have not understood very well how to find Laurent series of a-bit-harder-than-usual meromorphic function such as the one I proposed here… Could you tell me what I miss to compute this series?
Best Answer
What you did is fine. And you do not need to compute those Laurent series at all. For instance, at $1$, $\frac{\sin(z)}{z^3-z}$ is the quotient of a function which has not zero there by a function which has a simple zero there. Therefore $\frac{\sin(z)}{z^3-z}$ has a simple pole at $1$ and$$\operatorname{res}_{z=1}\left(\frac{\sin(z)}{z^3-z}\right)=\frac{\sin(1)}{3z^2-1|_{z=1}}=\frac{\sin(1)}2.$$