Write $\sin{x} = (e^{i x}-e^{-i x})/(2 i)$. Then consider the integral
$$PV \oint_{C_{\pm}} dx \frac{e^{\pm i z}}{z (z^2-2 z+2)} $$
where $C_{\pm}$ is a semicircular contour of radius $R$ in the upper/lower half plane with a semicircular detour into the upper/lower half plane of radius $\epsilon$. For $C_{+}$, we have
$$PV \oint_{C_{+}} dz \frac{e^{i z}}{z (z^2-2 z+2)} = \int_{-R}^{-\epsilon} dx \frac{e^{i x}}{x (x^2-2 x+2)}+ i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{e^{i \epsilon e^{i \phi}}}{\epsilon e^{i \phi} (\epsilon^2 e^{i 2 \phi} - 2 \epsilon e^{i \phi}+2)} \\+ \int_{\epsilon}^R dx \frac{e^{i x}}{x (x^2-2 x+2)}+ i R \int_0^{\pi} d\theta \, e^{i \theta} \frac{e^{i R e^{i \theta}}}{R e^{i \theta} (R^2 e^{i 2 \theta} - 2 R e^{i \theta}+2)} $$
For $C_-$, we have
$$PV \oint_{C_{-}} dz \frac{e^{-i z}}{z (z^2-2 z+2)} = \int_{-R}^{-\epsilon} dx \frac{e^{-i x}}{x (x^2-2 x+2)}+ i \epsilon \int_{-\pi}^0 d\phi \, e^{i \phi} \frac{e^{-i \epsilon e^{i \phi}}}{\epsilon e^{i \phi} (\epsilon^2 e^{i 2 \phi} - 2 \epsilon e^{i \phi}+2)} \\+ \int_{\epsilon}^R dx \frac{e^{-i x}}{x (x^2-2 x+2)}- i R \int_0^{\pi} d\theta \, e^{-i \theta} \frac{e^{-i R e^{-i \theta}}}{R e^{-i \theta} (R^2 e^{-i 2 \theta} - 2 R e^{-i \theta}+2)} $$
In both cases, we take the limits as $R \to \infty$ and $\epsilon \to 0$. Note that, in both cases, the respective fourth integrals have a magnitude bounded by
$$\frac{2}{R^2} \int_0^{\pi/2} d\theta \, e^{-R \sin{\theta}} \le \frac{2}{R^2} \int_0^{\pi/2} d\theta \, e^{-2 R \theta/\pi}\le \frac{\pi}{R^3}$$
The respective second integrals of $C_{\pm}$, on the other hand, become equal to $\mp i \frac{\pi}{2} $. Thus,
$$PV \oint_{C_{\pm}} dz \frac{e^{\pm i z}}{z (z^2-2 z+2)} = PV \int_{-\infty}^{\infty} dx \frac{e^{\pm i x}}{x (x^2-2 x+2)} \mp i \frac{\pi}{2}$$
On the other hand, the respective contour integrals are each equal to $\pm i 2 \pi$ times the sum of the residues of the poles inside their contours. (For $C_-$, there is a negative sign because the contour was traversed in a clockwise direction.) The poles of the denominator are at $z_{\pm}=1 \pm i$. Thus,
$$PV \int_{-\infty}^{\infty} dx \frac{e^{\pm i x}}{x (x^2-2 x+2)} \mp i \frac{\pi}{2} = \pm i 2 \pi \frac{e^{\pm i (1 \pm i)}}{(1 \pm i) (2) (\pm i)} $$
Taking the difference between the two results and dividing by $2 i$, we get that
$$\int_{-\infty}^{\infty} dx \frac{\sin{x}}{x (x^2-2 x+2)} = \frac{\pi}{2} \left (1+\frac{\sin{1}-\cos{1}}{e} \right ) $$
Note that we may drop the $PV$ because the difference between the integrals removes the pole at the origin.
On a little semicircle $\;C_\epsilon\;$ (upper) of radius $\;\epsilon>0\;$ about zero, you indeed have to take the limit $\;\epsilon\to0\;$ and then you get by the corollary to the lemma here , that
$$\lim_{\epsilon\to0}\int_{C_\epsilon}\frac{(z+1)^2}{z(z^2+4)^2}dz=\pi i\,\text{Res}\,(f)_{z=0}=\frac{\pi i}{16}$$
The rest is standard: the upper semicircle of radius $\;R>>0\;$ so that you only need the poles with positive imaginary part, take residues, limits and etc.
Added on request: The actual "problem" is to deal with the integral on the arc $\;\gamma_R:=\{z=Re^{it}\;/\;0<t<\pi\}\;$ , but either the L-M (Estimmation Lemma) or Jordan's Lemma solve that and this part of the integral, in this case, tends to zero when $\;R\to\infty\;$.
Also, the residue at $\;z=2i\;$, which is a double pole (observe that we don't care about the other pole $\;z=-2i\;$, as we are on the upper semiplane) is
$$\text{res}\,(f)_{z=2i}=\lim_{z\to 2i}\left((z-2i)^2\frac{(z+1)^2}{z(z^2+4)^2}\right)'=\lim_{z\to2i}\left(\frac{(z+1)^2}{z(z+2i)^2}\right)'$$$${}$$
$$=\lim_{z\to2i}\frac{2z(z+1)(z+2i)-(z+1)^2\left((z+2i)+2z\right)}{z^2(z+2i)^3}=\frac{4i(1+2i)\cdot4i-(1+2i)^2(4i+4i)}{(-4)(-64i)}=$$$${}$$
$$=\frac{-16-32i-(-3+4i)\cdot8i}{256i}=\frac{-16-32i+24i+32}{256i}=-\frac1{32}-\frac i{16}$$
so we get, calling $\;C\;$ the whole simple, closed contour:
$$-\frac1{32}-\frac i{16}=\oint_{C}f(z)dz=\int_{-R}^{-\epsilon}f(x)dx-\int_{\gamma_\epsilon}f(z)dz+\int_\epsilon^Rf(x)dx+\int_{\gamma_R}f(z)dz$$
Observe the minus sign before the integral around zero because when we go from $\;-R\to R\;$ that half semicircle's integration is done in the negative direction! Well, now just take the double limit and use Cauchy's Theorem:
$$\lim_{R\to\infty,\,\epsilon\to0}\left(-\frac1{32}-\frac i{16}\right)=\frac1{2\pi i}\lim_{R\to\infty,\,\epsilon\to0}\int_Cf(z)dz=\frac1{2\pi i}\left(\int_{-\infty}^\infty f(x)dx-\frac{\pi i}{16}\right)\implies$$$${}$$
$$\implies\int_{-\infty}^\infty\frac{(x+1)^2}{x(x^2+4)^2}dx=\frac\pi8 $$
Best Answer
$\omega=0$ is a double pole, therefore $$\lim_{\varepsilon\to0^+}\int_{C_\varepsilon}\frac{e^{i\alpha\omega}}{\omega^2}\mathrm{d}\omega \text{ is divergent}$$ which has also been shown by OP's equation (2). There is really no way to go around a double pole.
I guess the OP is trying to evaluate $$\int_{-\infty}^\infty\frac{e^{i\alpha z}}{z^2}\,\mathrm{d}z$$
One may try the Hadamard finite part (denoted as $\mathcal{H}$) which assigns a finite value for the singular integral $$ \mathcal{H}\int_{-\infty}^\infty\frac{e^{i\alpha z}}{(z-t)^2}\,\mathrm{d}z :=\frac{\mathrm{d}}{\mathrm{d}t}\left[\mathcal{P}\int_{-\infty}^\infty\frac{e^{i\alpha z}}{z-t}\,\mathrm{d}z\right] $$ where $\mathcal{P}$ denotes Cauchy principal value. Use the Sokhotski–Plemelj theorem to find $\mathcal{P}$, $$ \mathcal{P}\int_{-\infty}^\infty\frac{e^{i\alpha z}}{z-t}\,\mathrm{d}z = \frac{1}{2}\lim_{\epsilon\to0^+}\Bigg[\underbrace{\int_{-\infty}^\infty\frac{e^{i\alpha z}}{z-t- i\epsilon}\,\mathrm{d}z}_{=2\pi i e^{i\alpha(t+i\epsilon)}}+\underbrace{\int_{-\infty}^\infty\frac{e^{i\alpha z}}{z-t+ i\epsilon}\,\mathrm{d}z}_{=0}\Bigg] =\pi i e^{i\alpha t}\tag{*} $$ where the contour integration with upper semi-circle is used to get the integral values. Therefore, $$ \mathcal{H}\int_{-\infty}^\infty\frac{e^{i\alpha z}}{(z-t)^2}\,\mathrm{d}z =\frac{\mathrm{d}}{\mathrm{d}t}\left(\pi i e^{i\alpha t}\right)=-\alpha\pi e^{i\alpha t} $$ Letting $t=0$, $$ \mathcal{H}\int_{-\infty}^\infty\frac{e^{i\alpha z}}{z^2}\,\mathrm{d}z=-\alpha\pi $$
Warning: The Hadamard finite part got its name because it manually drops the divergent terms such as $\lim_{\varepsilon\to0^+}\int_{C_\varepsilon}e^{i\alpha\omega}/\omega^2\,\mathrm{d}\omega$ in this case as no one knows how to deal with the divergent terms. Therefore, one should always refer to the context of the (physical) problem to check if the $\mathcal{H}$ value makes any sense.
To reply @zytsang's question in the comment, suppose we know A Priori the result we want: $$\begin{cases} 0 & \quad \text{for}\quad\alpha>0 \\ 2\pi\alpha & \quad \text{for}\quad\alpha<0 \end{cases}$$ Denote $\mathcal{P}'_+$ and $\mathcal{P}'_-$ as the principal-values-as-we-want-it. The regularization in (*) can be manipulated as $$ \mathcal{P}'_+\int_{-\infty}^\infty\frac{e^{i\alpha z}}{z-t}\,\mathrm{d}z =\frac{1}{2}\lim_{\epsilon\to0^+}\Bigg[\underbrace{A\cdot\int_{-\infty}^\infty\frac{e^{i\alpha z}}{z-t- i\epsilon}\,\mathrm{d}z}_{=2\pi i A e^{i\alpha(t+i\epsilon)}}+\underbrace{B\cdot\int_{-\infty}^\infty\frac{e^{i\alpha z}}{z-t+ i\epsilon}\,\mathrm{d}z}_{=0}\Bigg] \quad \text{for }\alpha>0 \text{ (upper semi-circle)} $$ Setting $A=0$ gets the physical result for $\alpha>0$. Similarly, $$ \mathcal{P}'_-\int_{-\infty}^\infty\frac{e^{i\alpha z}}{z-t}\,\mathrm{d}z =\frac{1}{2}\lim_{\epsilon\to0^+}\Bigg[\underbrace{C\cdot\int_{-\infty}^\infty\frac{e^{i\alpha z}}{z-t- i\epsilon}\,\mathrm{d}z}_{=0}+\underbrace{D\cdot\int_{-\infty}^\infty\frac{e^{i\alpha z}}{z-t+ i\epsilon}\,\mathrm{d}z}_{=-2\pi i D e^{i\alpha(t+i\epsilon)}}\Bigg] \quad \text{for }\alpha<0 \text{ (lower semi-circle)} $$ Setting $D=2$ gets the physical result for $\alpha<0$. Because the original integral is divergent, if you are allowed to artificially specify the rate of taking the limit on different directions, you can get whatever you want.
Although Abel's quote is for divergent series, I think it is also relevant to the divergent integral. But I wouldn't mind the "shameful" part, though.