I came across this problem.
Let $C$ denote the circle {|z| =2}, parametrized as a positively oriented simple closed curve. Evaluate $\int_c\frac{1}{z^2-1}dz$
I want to approach this problem using the Cauchy Integral Formula. Let $g(z) = \frac{1}{z^2-1}$. $g(z)$ is analytic everywhere on the closed curve except at $z = \pm1.$ We can manipulate g(z) via partial fraction decomposition.
\begin{eqnarray} g(z) = \frac{1}{z^2-1}& = & \frac{1}{(z+1)(z-1)}\\ & = & \frac{-1/2}{z+1} + \frac{1/2}{z-1} \end{eqnarray} Thus we can have the following set up. \begin{eqnarray} \int_C\frac{dz}{z^2-1} & = & -\frac{1}{2}\int_C\frac{1}{z+1}dz + \frac{1}{2}\int_C\frac{1}{z-1}dz\\ & = & -\frac{1}{2}2\pi if_1(-1) + \frac{1}{2}2\pi if_2(1)\\ & = & -\pi i + \pi i = 0\\ \end{eqnarray}
Did I approach this correctly?
Thank you for your time and thank you for any feedback you provide.
The integral is indeed zero as you claim. This in fact can be seen easier since the integrand $g$ is an even function.
EDIT:
Allow me to expand on my comment. Suppose $f(z)$ is even, and $C$ is a circle centered around the origin, with an arbitrary radius $r>0$. We then have the simple parameterization for $C$: $$z(t)=r \exp{it};0 \leq t \leq 2 \pi, $$ and the integral is by definition $$\int_0^{2\pi} f(r \exp it)ir \exp it dt=\int_0^{\pi} f(r \exp it)ir \exp it dt+\int_\pi^{2\pi} f(r \exp it)ir \exp it dt $$
Using the substitution $t-\pi=\tau$ in the second integral gives $$\oint_C f(z)dz=\int_0^{\pi} f(r \exp it)ir \exp it dt+\int_0^\pi f(r \exp i(\tau+ \pi))ir \exp i(\tau+\pi)d\tau. $$ Simplifying the exponents, and using the fact that $f$ is even shows that the two sub-integrals cancel each other out.