Taking $\int_{-\infty}^{\infty} \frac{e^{i \omega x}}{1 + ix}dx$ through Residue Theory? In the text "Basic Complex Analysis" Third Edition by Jerrold E. Marsden and Michael J. Hoffman I'm inquiring if there's an alternate way through Complex-Analysis to evaluate $\text{Example (4.38)}$ ? For those who don't have the textbook on hand I've written the approach the authors have taken.
$\text{Example (4.3.8)}$
Let $\omega$ be a nonzero real constant and evaluate,
$$\int_{-\infty}^{\infty} \frac{e^{i \omega x}}{1 + ix}dx.$$
$\text{Solution}$
The author notes that this is an integral of the Fourier transform type with our choice of $f$ being $1/(1+iz)$. The author notes along the real axis one can note that
$(1)$
$$|g(x)| = |e^{-i \omega z} | / | 1 + ix| = 1 / \sqrt{1 + x^{2}}.$$
It's rather trivial to see that, 
$$ \int_{-\infty}^{\infty} | 1 / \sqrt{1 + x^{2}} | dx < \infty.$$
$\text{Commentary (1.0)}$
When the author noted, that the integral was of a Fourier Transform type it seems our problem boils down to evaluating the respective integrals in the form,
$(2)$
$$\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}f(z)e^{-iwz}dz$$
Or alternatively of the latter integrals of the form, 
$$\int_{-\infty}^{\infty}f(z)e^{-iwz}dz $$
$\text{Remark}$
An important detail one shouldn't miss is that our choice of $f$ is seen as $f(z) = q(z)/p(z)$
To obtain convergence of our integral, the author makes note that,
$(2)$
$$\int_{-\infty}^{\infty} \frac{e^{i \omega x}}{1 + ix}dx = \lim_{A \rightarrow \infty, B \rightarrow \infty} \int_{-A}^{B}\frac{e^{-i \omega x}}{1 + ix}dx.$$
$\text{Commentary (1.1)}$
This development rather is a key ingredient since convergence is a necessary condition also one can work out that, 
\begin{align*}
 \lim_{A \rightarrow \infty, B \rightarrow \infty} \int_{-A}^{B}\frac{e^{-i \omega x}}{1 + ix}dx&= \lim_{A \rightarrow \infty} \bigg( \int_{-A}^{B} f \, \bigg) dx \, +  \lim_{B \rightarrow \infty} \bigg( \int_{-A}^{B} f \bigg)dx\\
      &= \int_{-\infty}^{B} \big(f\big) \, dx + \int_{-A}^{\infty} \big(f) \, dx
\end{align*}
The author notes that since $|f(z)| = 1/|1+iz| \leq 1/ (|z|-1)$ for $|z| > 1$, this factor does not shrink towards $0$. For each $\epsilon > 0$,there is an $R(\epsilon)$ such that $|f(z)| < \epsilon$ whenever $|z| \geq R(\epsilon)$. Also the author notes the exponential factor will play nice on the half plane. But which plane in question depends on the sigh of $\omega$. If $z = x+iy$ with $x$ and $y$ real, then $|e^{i \omega z}| = e^{i \omega x + \omega y} = e^{wy}$. Therefore, $\omega < 0$ implies $|e^{i \omega z} | = e^{ \omega y} \leq 1 $ in the upper half plane $\mathcal{H.}$ In a  similar fashion one can say that,$\omega < 0$ implies $|e^{i \omega z} | = e^{ \omega y} \leq 1$ in the lower half plane $\mathcal{L.}$
$\text{Commentary (1.2)}$
Witnessing these ocean of developments it seems the location of the behavior depending on the sign of the exponential factor of our choice in $f$ in the planes $\mathcal{L}$ or $\mathcal{H.}$ Also as a consequence of this some of the analytic properties of our choice $f$ become restricted to either $\mathcal{L}$ or $\mathcal{H.}$ Immediate properties that come to mind are the locations of $\operatorname{Res}(f)$, where $f$ is analytic, etc. This intuition seems to be valid since the author calculates the residue's as follows 
$$\operatorname{Res}\bigg(\operatorname{\frac{e^{i \omega z}}{i(z-i)}; i}\bigg) \bigg|_{z = i} = \frac{e^{\omega}}{i}.$$ The furthermore sets that,
$$\Sigma_{\mathcal{H}} = \text{the sum of the residues in } \mathcal{H}= \frac{e^{\omega}}{i}$$
$$  \Sigma_{\mathcal{L}} = \text{the sum of the residues in } \mathcal{L}=0.$$ 
After dealing with the location of our Residue's considers that $A$ and $B$ are both larger that $R(\epsilon)$ and larger that $1$, we can consider the rectangular paths. The rectangular contours that are considered are displayed in $\text{Figure (4.3.2)}$. He then goes on to mention that $\gamma$ is the segment of the real axis from $-A$ to $B$. We have a two closed rectangular contours denoted by $\Gamma = \gamma + \mu_{1} + \mu_{2} + \mu_{3}$ counterclockwise through the upper half plane and as a rectangle $\wedge = \gamma + \nu_{1} + \nu_{2} + \nu_{3}.$ In each case the distance $C$ the real axis will be selected larger than $R(\epsilon)$ depend on $A$ and $B$. So one will have,
$(4)$
$$ \oint_{\Gamma} g = 2\pi i \Sigma_{\mathcal{H}} = 2 \pi e^{\omega} \text{and} \oint_{\wedge} g = -2\pi i \Sigma_{\mathcal{L}}= 0. $$
Thus,
$(5)$
$$\int_{-A}^{B} \frac{e^{i \omega x}}{1 + ix}dx = \oint_{\gamma} g = 2 \pi i \Sigma_{\mathcal{H}} \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \, \, \, \, \, \, \, \, \, -\bigg( \oint_{\mu_{1}} g + \oint_{\mu_{2}} g + \oint_{\mu_{3}} g\bigg) $$
$$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = -2 \pi i \Sigma_{\mathcal{L}\quad} \! \! \! \! - \bigg( \oint_{\nu_{1}}g + \oint_{\nu_{2}}g + \oint_{\nu_{2}}g \bigg)$$

$$\text{Figure (4.3.2): Paths for Example (4.3.8)}$$
$\text{Commentary (1.3)}$
The author supposes that $\omega < 0$, in the case that we have favorable behavior in $\mathcal{H}\quad$ Along $\mu_{1}$, also see that $z = B + iy$, so we have that
$(6)$
\begin{align*}
 \lim_{\epsilon \rightarrow \infty} \bigg| \oint_{\mu_{1}}g\bigg| &= \bigg| \int_{0}^{C} f(z)e^{i \omega z}dz\bigg| \\
      &= \bigg| \int_{0}^{C} f(-A+iy)e^{-i \omega(-A+iy)} idy\bigg| \leq \int_{0}^{C}\epsilon e^{\omega y}dy \leq \frac{r}{|\omega|} \rightarrow 0.
\end{align*} 
In order to get our desired bound over $\mu_{1}$, one has to note that,
$$\frac{e^{i \omega(B+iy)}}{1+iz} \leq \frac{i}{i}\bigg| \frac{e^{(i \omega \beta + i\omega y)}}{1 + z} \bigg| \leq \frac{i\epsilon}{i} \leq \epsilon.$$
The author notes a very similar process over $\mu_{3}$ hence,
$(7)$ 
\begin{align*}
 \lim_{\epsilon \rightarrow \infty} \bigg| \oint_{\mu_{3}}g\bigg| &= \bigg| \int_{C}^{0} f(z)e^{i \omega z}dz\bigg| \\
      &= \bigg| \int_{C}^{0} f(B+iy)e^{-i \omega(B+iy)} idy\bigg| \leq \int_{C}^{0}\epsilon e^{\omega y}dy \leq \frac{\epsilon}{|\omega|} \rightarrow 0.
\end{align*}
Now for the contribution over $\mu_{3}$ it's a very similar however some adjustments were made, the strict requirement is posed that $C > R(\epsilon)$ this is initially done since due to the location of $C$ it's contribution has to be small or else our estimates will be invalid. We have
$(8)$
\begin{align*}
 \lim_{\epsilon \rightarrow \infty} \bigg| \oint_{\mu_{3}}g\bigg| &= \bigg| \int_{B}^{-A} f(z)e^{i \omega z}dz\bigg| \\
      &= \bigg| \int_{-A}^{B} f(x+iC)e^{-i \omega(x+iC)} idy\bigg| \leq \int_{C}^{0}\epsilon e^{C \omega}dy \leq \epsilon(A+B)e^{\omega C} \rightarrow 0.
\end{align*}
$\text{Commentary $$(1.4)}$
The author remembers that $\omega > 0$, one can select $C$ such that $C > 1$, and also that $C > R(\epsilon)$ and large enough so that $(A + B)e^{- \omega C} < 1$. We find that for $A$ and $B$ larger then $(1)$ He discovers the bound by noting that $A$ and $B$ are respectively larger than $(1)$ and larger then $R(\epsilon)$ so finally we have,
$$\lim_{\epsilon \rightarrow \infty}\bigg| \int_{-A}^{B}g(x)dx + 2 \pi i \Sigma_{\mathcal{L}\quad} \bigg| \leq \sum_{\psi}  \bigg|\oint_{\nu_{\psi}} g\bigg| \leq  \bigg(\frac{2}{| \omega|} + 1 \bigg) \epsilon \rightarrow 0. $$
 A: What's being done by the author here seems—to me at least—to be over-complicated. We can evaluate this integral in a much simpler manner through application of Jordan's lemma.
To evaluate 
$$I(\omega)=\int_{-\infty}^\infty\frac{e^{i\omega x}}{1+ix}dx$$
we will consider a semicircular contour $\gamma_R$ of radius $R>1$ in the upper half of the complex plane like so:
         
             
             
             
          
We can think of our integral along $\gamma_R$ as the sum of two integrals
$$\oint_{\gamma_R}\frac{e^{i\omega z}}{1+iz}dz=\int_{C_R}\frac{e^{i\omega z}}{1+iz}dz+\int_{-R}^R\frac{e^{i\omega z}}{1+iz}dz\tag{1}$$
where $C_R$ is the curved portion of $\gamma_R$. If we take the limit $R\rightarrow\infty$ we see that 
$$\lim_{R\rightarrow\infty}\int_{-R}^R\frac{e^{i\omega z}}{1+iz}dz=I(\omega).$$
But this is not enough for our purposes, if we are to evaluate $I(\omega)$ by the residue theorem we must also have
$$\lim_{R\rightarrow\infty}\int_{C_R}\frac{e^{i\omega z}}{1+iz}dz=0\tag{2}$$
as we want the contribution along the arc to tend to $0$.
 
To show that $(2)$ happens we may apply Jordan's lemma. We consider $\omega>0$, then applying the lemma gives
$$\begin{align*}
\left | \int_{C_R}\frac{e^{i\omega z}}{1+iz}dz \right |&\leq\frac{\pi}{\omega}M_R\\
 &=\frac{\pi}{\omega}\cdot\underset{\theta\in[0,\,\pi]}{\text{max}}\left | \frac{1}{1+iRe^{i\theta}}\right |\\
 &=\frac{\pi}{\omega}\cdot\underset{\theta\in[0,\,\pi]}{\text{max}}\left | \frac{1-iRe^{-i\theta}}{1+R^2}\right |\\
&\leq \frac{\pi}{\omega}\cdot\frac{1+R}{1+R^2},\qquad\text{by the triangle inequality.}\tag{3}
\end{align*}$$
Since the expression $(3)$ goes to $0$ for all $\omega>0$ as $R$ increases, we can back-track through our manipulations and conclude the limit in $(2)$ holds true.
 
We may now turn to the integral on the left hand side of $(1)$ and apply the residue theorem. Remembering that we specified $R>1$ we have:
$$\begin{align*}
\oint_{\gamma_R}\frac{e^{i\omega z}}{1+iz}dz&=2\pi i\cdot\underset{z=i}{\text{Res}}\left[\frac{e^{i\omega z}}{1+iz}\right]\\
 &=\frac{2\pi}{e^{\omega}}.
\end{align*}$$
But the above holds for all $R>1$, so it holds in the limit, I.e.
$$\begin{align*}
\frac{2\pi}{e^{\omega}}&=\lim_{R\rightarrow\infty}\oint_{\gamma_R}\frac{e^{i\omega z}}{1+iz}dz\\
&=\lim_{R\rightarrow\infty}\underset{\text{goes to 0}}{\underbrace{\int_{C_R}\frac{e^{i\omega z}}{1+iz}dz}}+\lim_{R\rightarrow\infty}\int_{-R}^R\frac{e^{i\omega z}}{1+iz}dz\\
&=I(\omega).
\end{align*}$$
So
$$\int_{-\infty}^\infty\frac{e^{i\omega x}}{1+ix}dx=\frac{2\pi}{e^{\omega}}$$
for $\omega>0.$
