 Solution

Integrating from one to $R$ gives $(R^{\alpha +1}-1)/(\alpha +1)$ for $\alpha \ne -1$. It has a finite limit as $R\to \infty$ exactly when $\alpha +1<0$, while at $\alpha =-1$ the logarithm diverges. Nonnegativity equates finiteness of this integral with Lebesgue integrability. Hence $\boxed{x^{\alpha }\in L^1(1,\infty )⟺\alpha <-1}$.