First let's find the roots and factor $g(\omega)$ so the whole denominator of the integrand is factored completely. We'll need it factored to find residues anyway.
Completing the square to find the roots:
$$\begin{align*}g(\omega) = \alpha\omega^2 + i\beta\omega + \gamma &= 0\\
\\
\omega^2 + i\dfrac{\beta}{\alpha}\omega + \dfrac{\gamma}{\alpha}&= 0\\
\\
\omega^2 + i\dfrac{\beta}{\alpha}\omega - \left(\dfrac{\beta}{2\alpha}\right)^2 &= - \left(\dfrac{\beta}{2\alpha}\right)^2-\dfrac{\gamma}{\alpha}\\
\\
\left(\omega + i\dfrac{\beta}{2\alpha}\right)^2 &= - \left(\dfrac{\beta}{2\alpha}\right)^2-\dfrac{\gamma}{\alpha}\\
\\
\omega_{1,2} &= - i\dfrac{\beta}{2\alpha} \pm \sqrt{- \left(\dfrac{\beta}{2\alpha}\right)^2-\dfrac{\gamma}{\alpha}}\\
\\
\omega_{1,2} &= i\left[-\dfrac{\beta}{2\alpha} \pm \dfrac{\sqrt{ \beta^2+4\alpha\gamma}}{{2\alpha}}\right]\\
\\
\end{align*}$$
Thus
$$g(\omega)=\alpha\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)$$
and I'll stop using the $g(\omega)$ notation.
You are interested in integrals of the form
$$P.V. \int_{-\infty}^{\infty}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega$$
With $\alpha$, $t$, $\omega$, and $\omega_0$ real; and $\omega_1$ and $\omega_2$ complex.
I am specifically excluding cases where $\gamma$ is $0$, as that puts another singularity on the real axis (either $\omega_1$ or $\omega_2$ becomes 0) that needs to be handled specially. You can extend what I do later in this answer to handle that case, if you desire.
Consider the complex contour integral
$$\oint_{C}{\dfrac{ie^{iz t}}{\alpha\left(z-\omega_0\right)\left(z-\omega_1\right)\left(z-\omega_2\right)}}dz$$
This contour integral, the Residue Theorem, Jordan's Lemma, and a properly selected contour can be used to find the Principal Value of your improper integral on the real line.
You need to consider cases for $t >0$, $t<0$, and $t=0$, as they require different contours or different handling altogether.
For $t>0$, select closed contour $C_P$, that is a semi-circular contour in the upper half plane, that includes the real axis as its straight edge with a small semi-circular arc excursion into the upper half plane around $\omega_0$.
From the Residue Theorem
$$\begin{align*}&2\pi i \sum_{\omega_k \in \mathrm{UHP}} {\underset{z=\omega_k}{\mathrm{Res}}\left[\dfrac{ie^{iz t}}{\alpha\left(z-\omega_0\right)\left(z-\omega_1\right)\left(z-\omega_2\right)}\right] } = \oint_{C_P}{\dfrac{ie^{iz t}}{\alpha\left(z-\omega_0\right)\left(z-\omega_1\right)\left(z-\omega_2\right)}}dz\\
\\
&=\lim_{{r \to 0},{R \to \infty}}{\left[\int_{-R}^{\omega_0-r}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega +\\
\\
\int_{\omega_0+r}^{R}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega +\\
\\
\int_{\pi}^{0}{\dfrac{ie^{it\left(\omega_0+re^{i\omega}\right)}ire^{i\omega}}{\alpha\left(\omega_0+re^{i\omega}-\omega_0\right)\left(\omega_0+re^{i\omega}-\omega_1\right)\left(\omega_0+re^{i\omega}-\omega_2\right)}}d\omega +\\
\\
\int_{0}^{\pi}{\dfrac{ie^{itRe^{i\omega}}iRe^{i\omega}}{\alpha\left(Re^{i\omega}-\omega_0\right)\left(Re^{i\omega}-\omega_1\right)\left(Re^{i\omega}-\omega_2\right)}}d\omega\right]} \\
\\
&= P.V. \int_{-\infty}^{\infty}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega \space+\\
\\
&\lim_{r \to 0}{\left[ \dfrac{e^{i\omega_0 t}}{\alpha}\int_{0}^{\pi}{\dfrac{e^{itre^{i\omega}}}{\left(\omega_0+re^{i\omega}-\omega_1\right)\left(\omega_0+re^{i\omega}-\omega_2\right)}}d\omega \right]}+\\
\\
&\lim_{R \to \infty}{\left[\int_{0}^{\pi}{\dfrac{ie^{itRe^{i\omega}}iRe^{i\omega}}{\alpha\left(Re^{i\omega}-\omega_0\right)\left(Re^{i\omega}-\omega_1\right)\left(Re^{i\omega}-\omega_2\right)}}d\omega \right]}\\
\\
&= P.V. \int_{-\infty}^{\infty}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega \space+\dfrac{\pi e^{i\omega_0 t}}{\alpha\left(\omega_0-\omega_1\right)\left(\omega_0-\omega_2\right)}\space + 0\\
\end{align*}$$
So for $t>0$
$$\begin{align*}&P.V. \int_{-\infty}^{\infty}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega \\
\\
&= -\dfrac{\pi}{\alpha}\dfrac{ e^{i\omega_0 t}}{\left(\omega_0-\omega_1\right)\left(\omega_0-\omega_2\right)} - \dfrac{2\pi}{\alpha}\sum_{\omega_k \in \mathrm{UHP}} {\underset{z=\omega_k}{\mathrm{Res}}\left[\dfrac{e^{iz t}}{\left(z-\omega_0\right)\left(z-\omega_1\right)\left(z-\omega_2\right)}\right] }\\
\end{align*}$$
And, for example, for $t>0$, if both $\omega_1$ and $\omega_2$ are in the upper half plane, then working out the residues, one gets
$$\begin{align*}&P.V. \int_{-\infty}^{\infty}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega \\
\\
&= -\dfrac{\pi}{\alpha}\dfrac{ e^{i\omega_0 t}}{\left(\omega_0-\omega_1\right)\left(\omega_0-\omega_2\right)} - \dfrac{2\pi}{\alpha}\dfrac{ e^{i\omega_1 t}}{\left(\omega_1-\omega_0\right)\left(\omega_1-\omega_2\right)} - \dfrac{2\pi}{\alpha}\dfrac{ e^{i\omega_2 t}}{\left(\omega_2-\omega_0\right)\left(\omega_2-\omega_1\right)}\\
\end{align*}$$
\
For $t<0$, select closed contour $C_N$, that is a semi-circular contour in the lower half plane, that includes the real axis as its straight edge with a small semi-circular arc excursion into the lower half plane around $\omega_0$.
From the Residue Theorem
$$\begin{align*}&-2\pi i \sum_{\omega_k \in \mathrm{LHP}} {\underset{z=\omega_k}{\mathrm{Res}}\left[\dfrac{ie^{iz t}}{\alpha\left(z-\omega_0\right)\left(z-\omega_1\right)\left(z-\omega_2\right)}\right] } = \oint_{C_N}{\dfrac{ie^{iz t}}{\alpha\left(z-\omega_0\right)\left(z-\omega_1\right)\left(z-\omega_2\right)}}dz\\
\\
&=\lim_{{r \to 0},{R \to \infty}}{\left[\int_{-R}^{\omega_0-r}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega +\\
\\
\int_{\omega_0+r}^{R}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega +\\
\\
\int_{-\pi}^{0}{\dfrac{ie^{it\left(\omega_0+re^{i\omega}\right)}ire^{i\omega}}{\alpha\left(\omega_0+re^{i\omega}-\omega_0\right)\left(\omega_0+re^{i\omega}-\omega_1\right)\left(\omega_0+re^{i\omega}-\omega_2\right)}}d\omega +\\
\\
\int_{0}^{-\pi}{\dfrac{ie^{itRe^{i\omega}}iRe^{i\omega}}{\alpha\left(Re^{i\omega}-\omega_0\right)\left(Re^{i\omega}-\omega_1\right)\left(Re^{i\omega}-\omega_2\right)}}d\omega\right]} \\
\\
&= P.V. \int_{-\infty}^{\infty}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega \space+\\
\\
&\lim_{r \to 0}{\left[ -\dfrac{e^{i\omega_0 t}}{\alpha}\int_{-\pi}^{0}{\dfrac{e^{itre^{i\omega}}}{\left(\omega_0+re^{i\omega}-\omega_1\right)\left(\omega_0+re^{i\omega}-\omega_2\right)}}d\omega \right]}+\\
\\
&\lim_{R \to \infty}{\left[\int_{0}^{-\pi}{\dfrac{ie^{itRe^{i\omega}}iRe^{i\omega}}{\alpha\left(Re^{i\omega}-\omega_0\right)\left(Re^{i\omega}-\omega_1\right)\left(Re^{i\omega}-\omega_2\right)}}d\omega \right]}\\
\\
&= P.V. \int_{-\infty}^{\infty}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega \space-\dfrac{\pi e^{i\omega_0 t}}{\alpha\left(\omega_0-\omega_1\right)\left(\omega_0-\omega_2\right)}\space + 0\\
\end{align*}$$
So for $t<0$
$$\begin{align*}&P.V. \int_{-\infty}^{\infty}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega \\
\\
&= \dfrac{\pi}{\alpha}\dfrac{ e^{i\omega_0 t}}{\left(\omega_0-\omega_1\right)\left(\omega_0-\omega_2\right)} + \dfrac{2\pi}{\alpha}\sum_{\omega_k \in \mathrm{LHP}} {\underset{z=\omega_k}{\mathrm{Res}}\left[\dfrac{e^{iz t}}{\left(z-\omega_0\right)\left(z-\omega_1\right)\left(z-\omega_2\right)}\right] }\\
\end{align*}$$
And, for example, for $t<0$, if both $\omega_1$ and $\omega_2$ are in the lower half plane, then working out the residues, one gets
$$\begin{align*}&P.V. \int_{-\infty}^{\infty}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega \\
\\
&= \dfrac{\pi}{\alpha}\dfrac{ e^{i\omega_0 t}}{\left(\omega_0-\omega_1\right)\left(\omega_0-\omega_2\right)} + \dfrac{2\pi}{\alpha}\dfrac{ e^{i\omega_1 t}}{\left(\omega_1-\omega_0\right)\left(\omega_1-\omega_2\right)} + \dfrac{2\pi}{\alpha}\dfrac{ e^{i\omega_2 t}}{\left(\omega_2-\omega_0\right)\left(\omega_2-\omega_1\right)}\\
\end{align*}$$
Update 2
For $t=0$, we'll need to use the following result, which I won't prove here. But it can be proven using $u$ substitution, the definition of Cauchy Principal Value, and two different contour integrations that have closed contours that are infinitely long rectangles, with one edge along the ral axis with a infintely small semi-circular excursion around the pole on the real axis:
$$ P.V. \int_{-\infty}^\infty{\dfrac{1}{x-z_k}}dx = \pi i \space\mathrm{sgn}\left(\Im\left[z_k\right]\right)$$
We'll attack the $t=0$ case by performing Partial Fraction Expansion of the integrand and then applying the above result to resultant three simpler integrals
$$\begin{align*} & P.V. \int_{-\infty}^{\infty}{\dfrac{ie^{i\omega 0}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega \\
\\
&= P.V. \int_{-\infty}^{\infty}{\dfrac{i}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega \\
\\
&= P.V. \dfrac{i}{\alpha} \int_{-\infty}^\infty {\left[\dfrac{1}{\left(\omega_0-\omega_1\right)\left(\omega_0-\omega_2\right)\left(\omega-\omega_0\right)} \\
\\
+ \dfrac{1}{\left(\omega_1-\omega_0\right)\left(\omega_1-\omega_2\right)\left(\omega-\omega_1\right)} \\
\\
+ \dfrac{1}{\left(\omega_2-\omega_0\right)\left(\omega_2-\omega_1\right)\left(\omega-\omega_2\right)}\right]}d\omega\\
\\
&= \pi i\dfrac{i}{\alpha}\left[\dfrac{1}{\left(\omega_0-\omega_1\right)\left(\omega_0-\omega_2\right)}\mathrm{sgn}\left(\Im\left[\omega_0\right]\right)\\
\\
+ \dfrac{1}{\left(\omega_1-\omega_0\right)\left(\omega_1-\omega_2\right)}\mathrm{sgn}\left(\Im\left[\omega_1\right]\right)\\
\\
+ \dfrac{1}{\left(\omega_2-\omega_0\right)\left(\omega_2-\omega_1\right)}\mathrm{sgn}\left(\Im\left[\omega_2\right]\right)\right]\\
\end{align*}$$
\
Collecting all of the above results, we have
$$\begin{align*}&P.V. \int_{-\infty}^{\infty}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega \\
\\
&= 2\pi i\dfrac{i}{\alpha}\left[\dfrac{ e^{i\omega_0 t}}{\left(\omega_0-\omega_1\right)\left(\omega_0-\omega_2\right)}\dfrac{1}{2}\left[\mathrm{sgn}(t)+\mathrm{sgn}\left(\Im[\omega_0]\right)\right] + \\
\\
\dfrac{ e^{i\omega_1 t}}{\left(\omega_1-\omega_0\right)\left(\omega_1-\omega_2\right)} \dfrac{1}{2}\left[\mathrm{sgn}(t)+\mathrm{sgn}\left(\Im[\omega_1]\right)\right]+ \\
\\
\dfrac{ e^{i\omega_2 t}}{\left(\omega_2-\omega_0\right)\left(\omega_2-\omega_1\right)}\dfrac{1}{2}\left[\mathrm{sgn}(t)+\mathrm{sgn}\left(\Im[\omega_2]\right)\right]\right]\\
\end{align*}$$
Beardon's book is sadly out of print, but detailed proofs of results
concerning the winding number, or the index [incidentally, I
don't think that one can distinguish these two terms, at least not
in the way attempted in the question, although the same number
associated with a plane curve is defined in the literature in two
different ways, and the question is therefore an interesting one],
can also be found in Chapter 3 of Ash and Novinger,
Complex Variables (2004), which is downloadable
here.
Because of the importance of the concept, I think it is worth giving
a fairly short and straightforward proof from first principles,
rather than appealing to a series of general theorems (although that
is admittedly more efficient, and it would probably also be worth
giving a direct proof of the "fundamental theorem of calculus" for
rectifiable paths).
Digression:
This is especially so as it is not easy even to find books that
bother to treat the case of rectifiable curves. The best-known
exception may be Conway, Functions of One Complex Variable I
(2nd ed. 1978). Another exception is
Garling, A Course in Mathematical Analysis, Vol. III (2014).
But even these two books prove
the equality in question by using piecewise continuously
differentiable approximations to rectifiable curves; and so it is
hard, for me at least, to see the wood for the trees.
Rudin, in Principles of Mathematical Analysis (3rd ed. 1976),
remarks of the concept of rectifiability:
"The case $k = 2$ (i.e., the case of plane curves) is of
considerable importance in the study of analytic functions of a
complex variable."
In contrast, in Real and Complex Analysis (3rd ed. 1987),
the same author writes:
"In this section the required integration theory will be developed;
we shall keep it as simple as possible, and shall regard it merely
as a useful tool in the investigation of properties of holomorphic
functions. [...] A path is a piecewise continuously
differentiable curve in the plane."
It the latter, more utilitarian view of curves that seems to
dominate the literature on complex analysis. (There are some remarks
about this in Burckel, An Introduction to Classical Complex
Analysis, Vol. 1; but I haven't got that book yet, so I can't say
any more about it.)
(End of amateur lecture.) :)
Let $[a, b]$ be a compact interval of $\mathbb{R}$, and
$\gamma \colon [a, b] \to \mathbb{C}$ a continuous function. The
image set $[\gamma] = \gamma([a, b])$ is a compact subset of
$\mathbb{C}$, therefore it is closed.
Suppose $0 \notin [\gamma]$. Because $[\gamma]$ is closed, there
exists $R > 0$ such that the open disc $\mathbb{C}(0, R)$, with
centre $0$ and radius $R$, does not intersect $[\gamma]$.
Because $\gamma$ is a continuous function on a compact set, it is
uniformly continuous. So there exists a partition of $[a, b]$,
\begin{equation}
\tag{1}\label{eq:3257640:part:2}
a = t_0 < t_1 < \cdots < t_{n-1} < t_n = b,
\end{equation}
such that
$$
|\gamma(t) - \gamma(t')| < R \quad
(t_{j-1} \leqslant t < t' \leqslant t_j, \
1 \leqslant j \leqslant n).
$$
Therefore, each of the sets $\gamma([t_{j-1}, t_j])$ is contained in
an open disc (of radius $R$) not containing the origin. Because
there exists a continuous choice of logarithm on any such disc, a
simple inductive proof shows that
there exists a continuous function
$\lambda \colon [a, b] \to \mathbb{C}$ such that
$\gamma(t) = e^{\lambda(t)}$ for all $t \in [a, b]$.
Recall that $e^z = 1 + z + \rho(z)$, where $\rho(z)/z \to 0$ as
$z \to 0$.
Take $\epsilon > 0$ such that $|\rho(z)| < |z|/2$ when
$0 < |z| < \epsilon$. (For example, we could take
$\epsilon = \tfrac{3}{4}$. Not that it matters, of course!)
Because $\lambda$ is uniformly continuous, there exists
$\delta > 0$ such that
$|\lambda(t) - \lambda(t')| < \epsilon$ when $|t - t'| < \delta$.
Therefore,
\begin{align*}
|\gamma(t) - \gamma(t')| & = |e^{\lambda(t)} - e^{\lambda(t')}|
\\ & = |e^{\lambda(t')}||e^{\lambda(t) - \lambda(t')} - 1|
\\ & = |\gamma(t')||\lambda(t) - \lambda(t') +
\rho(\lambda(t) - \lambda(t'))|
\\ & \geqslant \frac{R|\lambda(t) - \lambda(t')|}{2}
\quad (|t - t'| < \delta).
\end{align*}
In a partition \eqref{eq:3257640:part:2} (not necessarily with the
same values of $t_j$ as before, of course),
if $|t_j - t_{j-1}| < \delta$ ($j = 1, \ldots, n$), then
$$
\sum_{j=1}^n|\lambda(t_j) - \lambda(t_{j-1})| \leqslant
\frac{2}{R}\sum_{j=1}^n|\gamma(t_j) - \gamma(t_{j-1})|.
$$
If $\gamma$ is rectifiable, and has length $L \geqslant 0$, it
follows that $\lambda$ is rectifiable, and has length at most
$2L/R$.
The path integral
\begin{equation}
\tag{2}\label{eq:3257640:path:2}
\int_\gamma \frac{dz}{z} = \int_a^b \frac{d\gamma(t)}{\gamma(t)}
\end{equation}
exists whenever the complex Riemann-Stieltjes integral on the right
is defined. (For a general definition of complex Riemann-Stieltjes
integrals, see e.g. Apostol, Mathematical Analysis (2nd ed. 1974).)
Given a partition \eqref{eq:3257640:part:2} - again, it has no
necessary relation to previous such partitions - "tagged", as they
say, with values $s_j \in [t_{j-1}, t_j]$ ($j = 1, \ldots, n$), the
Riemann-Stieltjes sum for the integral on the right of
\eqref{eq:3257640:path:2} is:
\begin{align}
& \phantom{={}}
\sum_{j=1}^n\frac{\gamma(t_j) - \gamma(t_{j-1})}{\gamma(s_j)}
\notag \\ & =
\sum_{j=1}^n\frac{e^{\lambda(t_j)} - e^{\lambda(t_{j-1})}}
{e^{\lambda(s_j)}}
\notag \\ & =
\sum_{j=1}^n\left(e^{\lambda(t_j) - \lambda(s_j)} -
e^{\lambda(t_{j-1}) - \lambda(s_j)}\right)
\notag \\ & =
\sum_{j=1}^n\left[\lambda(t_j) - \lambda(t_{j-1}) +
\rho(\lambda(t_j) - \lambda(s_j)) -
\rho(\lambda(t_{j-1}) - \lambda(s_j))\right]
\notag \\ & =
\tag{3}\label{eq:3257640:sum:log}
[\lambda(b) - \lambda(a)] +
\sum_{j=1}^n\rho(\lambda(t_j) - \lambda(s_j)) -
\sum_{j=1}^n\rho(\lambda(t_{j-1}) - \lambda(s_j)).
\end{align}
To prove that the Riemann-Stieltjes integral
\eqref{eq:3257640:path:2} exists, and that it is equal to
$\lambda(b) - \lambda(a)$, it is enough (see e.g. Exercise 7.3 of
Apostol) to prove that for every $\epsilon' > 0$ there exists
$\delta' > 0$ such that if $t_j - t_{j-1} < \delta'$
($j = 1, \ldots, n$), then
\begin{equation}
\tag{4}\label{eq:3257640:integral}
\left\lvert
\sum_{j=1}^n\frac{\gamma(t_j) - \gamma(t_{j-1})}{\gamma(s_j)} -
[\lambda(b) - \lambda(a)]
\right\rvert
< \epsilon'.
\end{equation}
If $L = 0$, i.e. if $\gamma$ is constant (a point "curve"), then
$\lambda$ is also constant, and \eqref{eq:3257640:integral} is
trivially satisfied; so, suppose from now on that $L > 0$; then at
least some of the "error" terms in \eqref{eq:3257640:sum:log} are
non-zero. Let $\eta > 0$ be such that
$$
\frac{|\rho(z)|}{|z|} < \frac{R\epsilon'}{2L}
\quad (0 < |z| < \eta).
$$
Let $\delta' > 0$ be such that
$$
|\lambda(t) - \lambda(t')| < \eta \quad (|t - t'| < \delta').
$$
Then, for any partition \eqref{eq:3257640:part:2} satisfying
$t_j - t_{j-1} < \delta'$ ($j = 1, \ldots, n$), tagged with values
$s_j \in [t_{j-1}, t_j]$ ($j = 1, \ldots, n$), the "error" in
\eqref{eq:3257640:sum:log} is bounded above, thus:
\begin{align*}
& \phantom{={}}
\left\lvert\sum_{j=1}^n\rho(\lambda(t_j) - \lambda(s_j)) -
\sum_{j=1}^n\rho(\lambda(t_{j-1}) - \lambda(s_j))\right\rvert
\\ & \leqslant
\sum_{j=1}^n\left(|\rho(\lambda(t_j) - \lambda(s_j))| +
|\rho(\lambda(t_{j-1}) - \lambda(s_j))|\right)
\\ & <
\frac{R\epsilon'}{2L}
\sum_{j=1}^n\left(|\lambda(t_j) - \lambda(s_j)| +
|\lambda(t_{j-1}) - \lambda(s_j)|\right)
\\ & \leqslant \epsilon'.
\end{align*}
We have thus proved
$$
\int_\gamma \frac{dz}{z} = \lambda(b) - \lambda(a),
$$
on the assumption only that $\gamma$ is rectifiable. If also
$\gamma$ is closed, i.e. $\gamma(a) = \gamma(b)$, then
$\operatorname{re}\lambda(a) = \operatorname{re}\lambda(b)$, and
$\operatorname{im}\lambda(t)$ is a choice of
$\operatorname{arg}\gamma(t)$ for $t \in [a, b]$, therefore:
$$
\int_\gamma \frac{dz}{z} = 2\pi{i}n(\gamma, 0),
$$
where I have used what seems to be the nearest thing to a standard
notation for the index, or winding number, of $\gamma$ about $0$.
Best Answer
In the paper, a function with 2 real variables is defined $$ I(x,y)=\frac1{2\pi i}\int_C\frac{F(u,v)}{u+iv-(x+iy)}d(u+iv). $$ Of this function, the Fourier transform in both variables is considered, it is assumed that the integration over the (compact) curve $C$ can be moved out of the Fourier integration \begin{align} \hat I(ω_x,ω_y)&=A_{Fourier}\int_{\Bbb R^2}\frac1{2\pi i}\int_C\frac{F(u,v)}{u+iv-(x+iy)}\,d(u+iv)\,e^{-i(xω_x+yω_y)}\,d(x,y) \\ &=\frac{A_{Fourier}}{2\pi i}\int_CF(u,v)\int_{\Bbb R^2}\frac{e^{-i(xω_x+yω_y)}}{u+iv-(x+iy)}\,d(x,y)\,d(u+iv) \\ &=\frac{A_{Fourier}}{2\pi i}\int_CF(u,v)e^{-i(uω_x+vω_y)}\int_{\Bbb R^2}\frac{e^{i((u-x)ω_x+(v-y)ω_y)}}{(u-x)+i(v-y)}\,d(x,y)\,d(u+iv) \\ &=\frac1{2\pi i}\int_CF(u,v)e^{-i(uω_x+vω_y)}\,d(u+iv)\cdot{A_{Fourier}}\int_{\Bbb R^2}\frac{e^{i(xω_x+yω_y)}}{x+iy}\,d(x,y) \end{align} For the last factor one now finds \begin{align} A_{Fourier}\int_{\Bbb R^2}\frac{e^{i(xω_x+yω_y)}}{x+iy}\,d(x,y) &={A_{Fourier}}\int_{\Bbb R^2}\frac{e^{i(xω_x+yω_y)}(ω_x-iω_y)}{(xω_x+yω_y)+i(yω_x-xω_y)}\,d(x,y) \\ &=A_{Fourier}\int_{\Bbb R^2}\frac{e^{i(xω_x+yω_y)}}{(xω_x+yω_y)+i(yω_x-xω_y)}\,\frac{d(xω_x+yω_y,yω_x-xω_y)}{ω_x+iω_y} \\ &=\frac{A_{Fourier}}{ω_x+iω_y}\int_{\Bbb R^2}\frac{e^{ix}}{x+iy}\,d(x,y) =\frac{A_{Fourier}}{ω_x+iω_y}\int_{\Bbb R^2}\frac{ix\sin(x)-iy\cos(x)}{x^2+y^2}\,d(x,y) \end{align} and somehow the last singular integral can be assigned a finite value.