Consider separately $\int_0^1 x^{-\alpha}\sin x\, dx \qquad\text{and}\qquad \int_1^\infty x^{-\alpha}\sin x\, dx.$
For the first integral, use the fact that for $0 we have $0\le \sin x \le x$. Now prove the convergence of the first integral by comparing our function with $\dfrac{1}{x^{\alpha-1}}$, noting that $\alpha-1<1$.
For the second integral, let $I(M)=\int_1^M x^{-\alpha} \sin x\,dx$ and examine the behaviour of $I(M)$ as $M\to\infty$.
There are various ways to proceed. One is to integrate by parts, letting $u=x^{-\alpha}$ and $dv=\sin x\,dx$. We get something which behaves nicely as $M$ gets large, and an integral of something that is a constant times $\dfrac{\cos x}{x^{1+\alpha}}$. This integral behaves nicely as $M\to\infty$, by comparison of $\dfrac{|\cos x|}{x^{1+\alpha}}$ with $\dfrac{1}{x^{1+\alpha}}$.
The reason for the integration by parts was to increase the power of $x$ at the bottom, in order to make the function decrease faster, fast enough for obvious convergence of the integral.