I want to compute the integral
$$\int_{|z – 1| = 1/2} \frac{e^{iz}}{(z^2-1)^2} \mathrm{d}z$$
using Cauchy's integral formula (residual theorem is not allowed). Examining the integrand one gets
$$ \frac{e^{iz}}{(z^2-1)^2} = \frac{e^{iz}}{(z-1)^2(z+1)^2}.$$
The problem is that I can't decompose the integrand as
$$f(z) \frac{1}{1-z}$$
since $f$ won't be holomorphic on $B_{1/2}(1)$ because the singularity of the integrand at $z=1$ is a pole of order $2$. I think a partial fraction decomposition won't help.
How should I proceed?
Best Answer
Recall Cauchy Integral Formula for Higher Derivatives: $f^{(n)}(z) = \frac{n!}{2\pi i} \displaystyle\int_\gamma \frac{f(\zeta)}{(\zeta-z)^{n+1}}\ d\zeta$.
Let $g(z)=\dfrac{e^{iz}}{(z+1)^2}$. Since $g$ is analytic on $\gamma: |z-1|=\frac{1}{2}$, $g'(1)=\frac{1}{2\pi i}\displaystyle\int_\gamma \frac{g(z)}{(z-1)^2}\ dz=\frac{1}{2\pi i}\displaystyle\int_\gamma \frac{e^{iz}}{(z^2-1)^2}\ dz$. Thus, the value of the integral is $2\pi i\cdot g'(1)$.