I assume there is a typo and you meant $$f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty}\frac{2i \sin(\pi \omega)}{\omega^{2}-1} \, e^{i t \omega} \, d {\color{red}{\omega}}. \tag{1} $$
And I assume the definition of the Fourier transform you're using is $$\mathcal{F}[f(t)](\omega) = \int_{-\infty}^{\infty} f(t) e^{-i \omega t} \, dt. $$
We can express $(1)$ as the difference of two Cauchy principal value integrals.
Specifically, $$ \begin{align} f(t) &= \frac{1}{2\pi} \int_{-\infty}^{\infty}\frac{2i \sin(\pi \omega)}{\omega^{2}-1} \, e^{i t \omega} \, d {\color{red}{\omega}} \\ &= \frac{1}{2 \pi} \int_{-\infty}^{\infty} \frac{e^{i \pi \omega} - e^{- i \pi \omega}}{w^{2}-1} \, e^{i t \omega } \, d \omega \\ &= \frac{1}{2 \pi} \left( \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{i \omega(t+ \pi)}}{w^{2}-1} \, d \omega - \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{i \omega(t- \pi)}}{w^{2}-1} \, d \omega \right). \end{align}$$
If $a \ge 0$, the magnitude of the function $e^{iaz} $ is bounded in the upper half-plane.
And if $a \le 0$, the magnitude of $e^{iaz}$ is bounded in the lower half-plane.
So if ${\color{red}{t \ge - \pi}}$, we can evaluate the first integral by integrating the function $$h(z) = \frac{e^{i z(t+ \pi)}}{z^{2}-1}$$ around a closed semicircular contour in the upper half-plane that has small half-circle indentations around the simple poles at $z=-1$ and $z=1$.
Letting the radii of the indentations go to zero and the radius of the big arc go to infinity, we get $$ \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{i \omega(t+ \pi)}}{w^{2}-1} \, d \omega - i \pi \, \text{Res} [h(z), -1] - \, i \pi \, \text{Res} [h(z), 1] =0,$$
which implies that $$ \begin{align} \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{i \omega(t+ \pi)}}{w^{2}-1} \, d \omega &= i \pi \, \text{Res} [h(z), 1] + \, i \pi \, \text{Res} [h(z), -1] \\ &= i \pi \left( \frac{e^{i (t+ \pi)}}{2} - \frac{e^{-i (t+ \pi)}}{2} \right) \\ &= - \pi \sin (t+ \pi) \\ &=\pi \sin t. \end{align}$$
I used fact that if a function $f(z)$ has a simple pole at $z_{0}$, then $$\lim_{r \to 0} \int_{C_{r}} f(z) \, dz = i \alpha \, \text{Res}[f(z), z_{0}],$$ where $C_{r}$ is an arc of the circle $|z- z_{0}|=r$ of angle $\alpha$. (This is sometimes referred to as the fractional residue theorem.)
And if ${\color{red}{ t \le \pi}}$, we can evaluate the second integral by integrating the function $$ g(z) = \frac{e^{i z(t- \pi)}}{z^{2}-1}$$ around a similar contour in the lower half-plane.
This causes us to detour around the poles in the opposite direction (i.e., counterclockwise.)
We end up with $$\begin{align} \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{i \omega(t- \pi)}}{w^{2}-1} \, d \omega &= {\color{red}{-}}i \pi \, \text{Res} [g(z), 1] {\color{red}{-}}i \pi \, \text{Res} [g(z), -1] \\ &= -i \pi \left( \frac{e^{i (t- \pi)}}{2} - \frac{e^{-i (t- \pi)}}{2} \right) \\ &= \pi \sin (t- \pi) \\ &=-\pi \sin t. \end{align}$$
So if ${\color{red}{|t| \le \pi}}$, $$\frac{1}{2\pi} \int_{-\infty}^{\infty}\frac{2i \sin(\pi \omega)}{\omega^{2}-1} \, e^{i t \omega} \, d \omega= \frac{1}{2 \pi} \left(\pi \sin t + \pi \sin t \right) = \sin t.$$
If $t< - \pi$, both $h(z)$ and $g(z)$ have to be integrated in the lower-half plane.
And if $t > \pi$, both $h(z)$ and $g(z)$ have to be integrated in the upper-half plane.
In both cases, this results in $f(t)=0$.
