$x^p$ is integrable near $0$ exactly for $p>-1$, and near $\infty$ exactly for $p<-1$.Suggest a slogan
Let \(p\in\RR\) and let \(f(x)\coloneqq x^{p}\) for \(x>0\).
\(f\in L^1((0, 1))\) if and only if \(p > -1\).
\(f\in L^1((1, \infty))\) if and only if \(p < -1\).
For \(0<a<b\), \(\int_a^b x^{p}\dx\) equals \(\frac{b^{p+1} - a^{p+1}}{p+1}\) if \(p\neq -1\) and \(\log b - \log a\) if \(p = -1\). By monotone convergence, \(\int_{(0,1)}x^{p}\dx = \lim_{a\to 0^+}\int_a^1 x^{p}\dx\), which is finite exactly when \(p+1>0\), and \(\int_{(1,\infty)}x^{p}\dx = \lim_{b\to\infty}\int_1^b x^{p}\dx\), which is finite exactly when \(p+1<0\).
Where is $x^p$ integrable in $\RR$? (Depending on $p$)