The threshold is $p=1$: below it $x^{-p}$ is integrable at $0$, above it at $\infty$.Suggest a slogan
For \(p\in\RR\), \[ \begin{aligned} \int_0^1 \frac{dx}{x^p} < \infty &\iff p < 1, \\ \int_1^\infty \frac{dx}{x^p} < \infty &\iff p > 1. \end{aligned} \]
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 \(p = 1\). Letting \(a\to 0^+\) with \(b = 1\) gives a finite limit exactly when \(1-p>0\); letting \(b\to\infty\) with \(a = 1\) gives a finite limit exactly when \(1-p<0\).
For which $p$ do $\int_0^1 x^{-p}$ and $\int_1^\infty x^{-p}$ converge?