Is $ G(x)=\int^\infty_{-\infty} \frac{e^{ix\omega}}{1+|\omega|^\alpha}d\omega$ an $L^1$ function?

While I am dealing with the non-local PDE problem$$((-\Delta)^{s}+I)u=f\qquad\text{in }\mathbb R^n$$I came across the following integral when I tried to compute its Green's function

$$G(x)=\int^\infty_{-\infty} \frac{e^{ix\omega}}{1+|\omega|^\alpha}d\omega$$where $\alpha\in(0,2)$.

Question: Is it true that $\displaystyle\int^\infty_{-\infty}|G(x)|dx<\infty$ ?

What is known:

• Note that $G(x)$ diverges at $x=0$, but is well-defined everywhere else with conditional convergence ($\alpha\in(0,1]$) or absolute convergence ($\alpha\in(1,2)$).

• If $\alpha=2$, $G(x)=\frac12 e^{-|x|}$. In general, for $\alpha$ taking integer values $\ge 2$, the closed form of the integral can be found using residue calculus.

• A similar integral is mentioned in a Math Overflow post where a user stated in a comment that $\displaystyle G(x)\approx\min(|x|^{-n-\alpha},|x|^{-n+\alpha})$, where in our case $n=1$. If this is true, $G$ would certainly be absolutely integrable. However, I do not know how to prove the stated asymptotics.