I'm trying to see for which values of a does the integral
$$ \int _0^{\infty }\:\frac{\sin\left(x\right)}{\left(1+x\right)^a} $$
converge.
I know (from this answer: For what values of $\alpha$, does this integral converge?) that for $$1\le a<2$$ the integral
$$
\int _0^{\infty }\:\frac{\sin\left(x\right)}{\left(x\right)^a}$$
converges, but didn't find it very useful...
The function $$\frac{\sin x}{(1+x)^\alpha}$$ is well defined at $0$ and its values is $0$. Hence you must check the behaviour at infinity.
Since $$\frac{\sin x}{(1+x)^\alpha} \sim\frac{\sin x}{x^\alpha}$$ when $x$ runs to infinity, your case is very similar to the one here: For what values of $\alpha$, does this integral converge?
Note that $$\int_0^\infty \frac{\sin x}{(1+x)^\alpha}dx \le \int_0^\infty \frac{1}{(1+x)^\alpha}dx$$
Take $y=x+1$, then $dx=dy$ and your integral becomes $$\int_1^\infty \frac{dy}{y^\alpha}$$ which converges if and only if $\alpha >1$.
If $\alpha<1$ then your integral becomes
$$\int_0^\infty (1+x)^{-\alpha}\sin x dx$$ and it clearly diverges. It remains only to deal with the case $\alpha=1$. In such case the integral converges to an improper Riemann integral.