In short, the reason for this particular choice of an interval is condition (iii). What you said about taking any $(a\pi,(a+2)\pi)$ is not entirely correct (at least if you want the principal branch of the logarithm), because if you take any $a$ that is not an odd integer, your interval will contain a multiple of $\pi$ of the form $(2n+1)\pi$ with $n\in \mathbb{Z}.$ But such an argument would mean that the associated complex number lies on the negative real axis which we removed from the domain $\Omega.$ Hence, the only options for the interval are $((2k+1)\pi,(2k+3)\pi)$ with $k\in\mathbb{Z}.$
In principle any $k$ is sufficient to fulfill conditions (i) and (ii), but let us now also consider condition (iii). Take any $k$ other than $-1,$ then the interval for $\theta$ will not contain $0.$ Therefore, the corresponding branch of the logarithm can never take real values so it cannot fulfill condition (iii). We are therefore left with only this particular choice of interval.
You can apply similar reasoning to any domain $\Omega$ of the form
$$\mathbb{C}\setminus \{re^{i\varphi}:r\in \mathbb{R},r\geq 0\}$$ where you take the entire complex plane and remove just the ray starting at the origin with angle $\varphi\in\mathbb{R}.$ As an exercise, you can think about which choices for $\phi$ are allowed to fulfill the conditions of Theorem 6.1. You could also remove any other curve starting at the origin and going to infinity (like an Archimedian or logarithmic spiral) but then your argument in the formula of the logarithm will also depend on $r.$
The limit $R \rightarrow \infty$ is irrelevant.
Inside the integral, $r$ is not the variable of integration nor is it in the integration bounds, so under certain conditions, the limit of $r \rightarrow 0$ and the integration operation can be exchanged.
So can you try making the substitutions
$$r = \dfrac{1}{n}$$
$$f_n(\omega) = \frac{e^{i\frac{1}{n}e^{i\omega}t}}{\left(\frac{e^{i\omega}}{n}+\omega_0-\omega_1\right)\left( \frac{e^{i\omega}}{n}+\omega_0 -\omega_2\right)}$$
$$f(\omega) = \frac{1}{\left(\omega_0 -\omega_1\right)\left(\omega_0 -\omega_2\right)}$$
$$|f_n(\omega)| = \frac{1}{\left|\frac{e^{i\omega}}{n}+\omega_0-\omega_1\right|\left| \frac{e^{i\omega}}{n}+\omega_0 -\omega_2\right|}$$
assuming
$$n > \dfrac{1}{\min\left(|\omega_0-\omega_1|,|\omega_0-\omega_2| \right)}$$
and try applying the Dominated Convergence Theorem?
Namely, find an integrable function $g(\omega)$ such that
$$|f_n(\omega)| \le g(\omega)$$
for all finite real $\omega_0$ and all finite, distinct, complex $\omega_1$, $\omega_2$. (For any specific choice of distinct $\omega_0$, $\omega_1$, and $\omega_2$, it should be straightforward to select a suitable $g(\omega)$.)
I didn't, and still haven't, done this rigorously myself.
I'm thinking the constant function
$$g(\omega) = \dfrac{1}{\left[\min\left(|\omega_0-\omega_1|,|\omega_0-\omega_2| \right) - \dfrac{1}{n_{min}}\right]^2}$$
might work.
Update
In my original answer, for "Term 3", instead of parameterizing the small semi-circular contour around $\omega_0$ and going through the above gyrations, I could have left "Term 3" as
$$\lim_{r \to 0} \int_{C_r}{\dfrac{ie^{iz t}}{\alpha\left(z-\omega_0\right)\left(z-\omega_1\right)\left(z-\omega_2\right)}}dz$$
and then used the lemma in this answer.
Knowing that the clockwise orientation of the contour simply changes the sign of the answer, one can obtain
$$\lim_{r \to 0} \int_{C_r}{\dfrac{ie^{iz t}}{\alpha\left(z-\omega_0\right)\left(z-\omega_1\right)\left(z-\omega_2\right)}} dz = {\dfrac{\pi e^{i\omega_0 t}}{\alpha\left(\omega_0-\omega_1\right)\left(\omega_0-\omega_2\right)}}$$
right away.
Best Answer
I think your first question is, why is there no continuous logarithm function defined in the complex plane, omitting $0$. The answer is, trace out a circle of radius $1$ centered at the origin, starting at $1$, and evaluate the logarithm function as you go around. Go ahead, do it! You will find that you either make a discontinuity somewhere along the way, or else when you get back to $z=1$ you are off by $2\pi i$ from the value you started with.
Now the same thing happens with the argument function. Start it with, say, the value $0$ at $z=1$, go around that circle (counterclockwise, just to be specific), and when you come back to $z=1$ the argument will be $2\pi$; discontinuity! The cure for the discontinuity is to make it impossible to go all the way around the circle. The easiest way to do that is to remove some ray that starts at the origin and goes to infinity. If the ray you remove is the non-positive reals, then the argument, starting with the value $0$ at $z=1$, will increase almost but not quite to $\pi$ as you go around counterclockwise (not reaching the excised ray), and will decrease almost but not quite to $-\pi$ as you go around the other way.
The non-positive imaginary part ray has argument $3\pi/2$ as you approach it counterclockwise from $z=1$, and argument $-\pi/2$ as you approach it the other way, so if you omit that ray then the argument can be taken to be between $-\pi/2$ and $3\pi/2$.
If I have missed the point of your question, please try again.