Problem: Analyze convergence of an improper integral $I=\int_0^\infty \frac{\sin{x}-x\cos{x}}{x^\alpha}dx$.
My work: Problematic points are $0$ and $\infty$. Therefore, we will write integral as a sum of two integrals : $$I=\int_0^1 \frac{\sin{x}-x\cos{x}}{x^\alpha}dx+\int_1^\infty \frac{\sin{x}-x\cos{x}}{x^\alpha}dx=I_1+I_2.$$ Second integral converges absolutely for $\alpha>3$. First integral converges for $\alpha<2$ by comparison test ($I_1\leq \int_0^1 \frac{dx}{x^{\alpha-1}}$).
Now I'm stuck at this point of problem. Any hint or help is welcome. Thanks in advance.