The problem I was asked to solve is the following:
Let $R > 1$, and considere $\Omega = \{z; |z| < R$ and $0 < \Re(z)\}$ and $\gamma = \partial \Omega$. Calculate $$\int_{\gamma} \frac{e^{iz}}{z^2 + 1}$$
I thought avoiding the singularities with semicircles of radius $\epsilon$ and then looking at limit with $\epsilon$ going to zero would be a good aproach, but I'm having some problems with that:
First of all, if $f(z) = \frac{e^{iz}}{z^2 + 1}$, then the integral $\int_{\gamma} f(z) = \int_0^1 f(\gamma(t))\gamma'(t)dt$ is not really defined. How do I know this limit with $\ \epsilon \to 0$ actually has something to do with this integral? Could be something completely unrelated.
I was using the following curves, all with $t \in [0,1]$:
- $\gamma_0(t) = Re^{i\pi t}$
- $\gamma_1(t) = Rit + (i+\epsilon)(1-t)$
- $\gamma_2(t) = i + \epsilon e^{i\pi t}$
- $\gamma_3(t) = (i - \epsilon)t + (-i + \epsilon)(1-t)$
- $\gamma_4(t) = -i + \epsilon e^{i\pi t}$
- $\gamma_5(t) = (-i + \epsilon)t -R(1-t)$
But I can't seem to be able to solve the integrals $\int_0^1 f(\gamma(t))\gamma'(t)dt$ with these parametrizations. Is this a poor aproach? Would you suggest something different?
In case you're familiar with Cauchy's Integral Formula, consider the following approach: It is $\frac{e^{iz}}{z^2 + 1} = \frac{e^{iz}}{(z-i)(z+i)} = \frac{\frac{e^{iz}}{z-i}}{z-(-i)}$; Define $g(z) := \frac{e^{iz}}{z-i}$, and now use Cauchy's Integral Formula (CIF): $\int \limits_\gamma \frac{e^{iz}}{z^2 + 1} \mathrm{d}z = \int \limits_\gamma \frac{g(z)}{z-(-i)} \mathrm{d}z \overset{\text{CIF}}{=} 2 \pi i \cdot g(-i) = 2 \pi i \cdot \frac{e^{i(-i)}}{(-i) - i} = -e \pi$. You may use the CIF here because $g$ is holomorphic inside and on the curve $\gamma$ (due to $R > 1$).