In the text "Basic Complex Analysis" Third Edition by Hoffman, and Marden I'm trying to proof $(i) - (iii)$, so far I've managed to execute on a attack towards the proof of $(iii)$ is the progress I've made valid, and also are there any ways to improve on the results I've brought to the table ?
$ \text{Proposition (4.3.9)}$ Under either of the situations $(i)$ or $(ii)$ described below, the improper integral $\int_{-\infty}^{\infty} f(x)e^{i \omega x} dx$ convergences to the value given by the corresponding formula. If $f(x)$ is real for real $x$, then the integrals $\int_{-\infty}^{\infty}f(x)cos(\omega x)dx $ and $\int_{-\infty}^{\infty} sin(\omega x) dx$ are equal respectively to its real part and the negative of its imaginary part
$(i)$ $\omega < 0:$ Suppose that $f$ is analytic on an open set containing the closed upper half plane $\mathcal{H}$ except for a finite number of isolated singularities none of which are on the real axis. Suppose also that $f(z) \rightarrow 0$ as $z \rightarrow \infty$ in that half plane in the sense that for each $\epsilon > 0$ there is an $R(\epsilon)$ such that $|f(z)| < \epsilon $ whenever $|z| \geq R(\epsilon)$ and $z\in \mathcal{H}$. Then one can observe that
$$ \int_{-\infty}^{\infty}f(x)e^{-i\omega x}dx = 2 \pi i\mathcal{\Sigma} \big\{ \text{residues of } f(z)e^{-i \omega z} \big\} \, \text{in} \, \mathcal{H}.$$
$(ii)$ $\omega > 0:$ If the conditions of $(i)$ hold with $\mathcal{H}$ replaced by the closed lower half plane $\mathcal{L}$, then one can observe that
$$\int_{-\infty}^{\infty}f(x)e^{-i\omega x}dx = 2 \pi i\mathcal{\Sigma} \big\{ \text{residues of } f(z)e^{-i \omega z} \big\} \, \text{in} \, \mathcal{L}. $$
$(iii)$ Both $(i)$ and $(ii)$ are vaild if $f = P/Q$ where $P$ and $Q$ are polynomials. The degree of $Q >P$, and $Q$ has no zeros on the real axis
$\text{Proof of } (i)$:
Assuming that $R > 1$, one can define $\gamma_{R}$ on $\mathcal{H}$ such that $$\gamma_{R}^{1}(t) = t + i0 \, \, \text{if} \, \, -R \leq t \leq R$$ $$\gamma_{R}^{2}(t) = Re^{it} \, \, \, \text{if} \, \, \, \, \, \, 0\leq t \leq \pi.$$
After settling the business of defining $\gamma_{R} \subset \mathcal{H}$, now we assume that $\mathcal{deg}(Q(z)) > \mathcal{deg}(P(z))$ as well our choice of $f$ being $\frac{p(z)}{q(z)}e^{i \omega z}$. Now we have to consider over $\gamma_{R}$ that,
$$\! \! \! \! \! \! \! \! \! \! \! \! \oint_{\gamma_{R_{\mathcal{H}}}} \frac{p(z)}{q(z)}e^{i \omega z}dz = 2 \pi i \bigg( \sum \mathcal{Res_{f}(P_{j}}) \cdot \operatorname{Ind_{\gamma}}(P_{j}) \bigg) $$
$$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,\, \, \, \, \, \, \, \,\, \, \, \, \, \, \, \, \, = 2 \pi i \bigg[ \sum_{j=1}^k\,\frac{1}{\left(m_j-1\right)!}\,\lim_{z\to z_j}\,\frac{\text{d}^{m_j-1}}{\text{d}z^{m_j-1}}\,\left(\frac{\left(z-z_j\right)^{m_j}\,p(z)}{p(z)}\right) \bigg]$$ $$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = 2 \pi i\mathcal{\Sigma} \big\{ \text{residues of } f(z)e^{-i \omega z} \big\} \, \text{in} \, \mathcal{H}. $$
But this isn't enough to realize $(i)$, through Cauchy's theorem and the configuration of $\gamma_{R_{\mathcal{H}}}$ one can notice that,
$$\oint_{\gamma_{R_{\mathcal{H}}}} \frac{p(z)}{q(z)}e^{i \omega z}dz = \oint_{\gamma_{R^1_{\mathcal{H}}}}\frac{p(z)}{q(z)}e^{i \omega z}dz + \oint_{\gamma_{R^{2}_{\mathcal{H}}}}\frac{p(z)}{q(z)}e^{i \omega z}dz.$$
It's worthwhile to claim that,
$$\oint_{\gamma_{R^{1}_{\mathcal{H}}}} \frac{p(z)}{q(z)}e^{i \omega z}dz \rightarrow \int_{{\, - \infty }}^{{\,c}}{{\frac{p(x)}{q(x)}e^{i \omega x}\,dx}} + \int_{{\, c}}^{{\,\infty}}{{\frac{p(x)}{q(x)}e^{i \omega x}\,dx}} \, \text{as} \, R \rightarrow \infty $$
Before there's any panic it can be noted that $p(z) = b_{m}z^{m}+ \cdot \cdot \cdot+b_{1}z+b_{o}$ is bounded for $|z| > R$ by repeated application of the triangle inequality:
$$|p(z)| \leq |b_{m}z^{m}| + \cdot \cdot \cdot |b_{1}z| + |b_{o}| \leq |b_{m}|R^{m} + \cdot \cdot \cdot + |b_{1}|R + |b_{o}|$$
Since $|z| > R$, we can finally say
$$|f(z)| = \frac{|f(z)||e^{i \omega z}|}{|q(z)|} \leq \frac{|b_{m}|R^{m} + \cdot \cdot \cdot + |b_{1}|R + |b_{o}|}{|a_{n}|R^{n}/2} \leq \frac{M}{R^{n-m}}$$
Now finally we obtain the bound,
$$\bigg| \oint_{\gamma_{R^{2}_{\mathcal{H}}}} \frac{p(z)}{q(z)}e^{i \omega z}dz \bigg| \leq \frac{M(2 \pi R}{R^{n-m}} = \frac{2M \pi}{R^{n-m-1}}.$$
Finally,
$$\lim_{R \rightarrow \infty} \bigg| \oint_{\gamma_{R^{2}_{\mathcal{H}}}} \frac{p(z)}{q(z)}e^{i \omega z}dz \bigg| \rightarrow 0 $$
Now to finish off $(i)$, we put the following together,
$$ \lim_{R \rightarrow \infty} \oint_{\gamma_{R_{\mathcal{H}}}} \frac{p(z)}{q(z)}e^{i \omega z}dz = \lim_{R \rightarrow \infty}\oint_{\gamma_{R^{1}_{\mathcal{H}}}} \frac{p(z)}{q(z)}e^{i \omega z}dz + \oint_{\gamma_{R^{2}_{\mathcal{H}}}} \frac{p(z)}{q(z)}e^{i \omega z}dz = \int_{-\infty}^{\infty}\frac{p(x)}{q(x)}e^{i \omega z}dx = 2 \pi i \bigg[ \sum_{j=1}^k\,\frac{1}{\left(m_j-1\right)!}\,\lim_{z\to z_j}\,\frac{\text{d}^{m_j-1}}{\text{d}z^{m_j-1}}\,\left(\frac{\left(z-z_j\right)^{m_j}\,p(z)}{p(z)}\right) \bigg]$$