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

Solution

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

 Solution (source code)