$x^{-p}$ 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 \(\frac{b^{1-p} - a^{1-p}}{1-p}\) 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 \(1-p>0\), and \(\int_{(1,\infty)}x^{-p}\dx = \lim_{b\to\infty}\int_1^b x^{-p}\dx\), which is finite exactly when \(1-p<0\).
Where is ${1\over x^p}$ integrable in $\RR$? (Depending on $p$)