Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / D / D3 / i

The function is nonnegative and measurable. For $\alpha \ne -1$, integrate from $\epsilon$ to one to get $(1-{\epsilon }^{\alpha +1})/(\alpha +1)$. The limit is finite exactly when $\alpha +1>0$. At $\alpha =-1$ the integral is $-\log \epsilon \to \infty$. Thus $\boxed{x^{\alpha }\in L^1(0,1)⟺\alpha >-1}$.

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 $\boxed{x^{\alpha }\in L^1(1,\infty )⟺\alpha <-1}$.