When do integrals of powers converge?

At infinity. For $p\ne1$ and $R>1$,
$$\int_1^R\frac{dx}{x^p}=\frac{R^{1-p}-1}{1-p}.$$
- If $p>1$, $R^{1-p}\to0$ and the integral converges to $\frac1{p-1}$.
- If $p<1$, $R^{1-p}\to+\infty$: divergence.
- If $p=1$: $\int_1^R\frac{dx}x=\ln R\to+\infty$: divergence.

So $\int_1^{+\infty}x^{-p}dx$ converges iff $p>1$.

At zero. For $p\ne1$ and $0<\varepsilon<1$,
$$\int_\varepsilon^1\frac{dx}{x^p}=\frac{1-\varepsilon^{1-p}}{1-p}.$$
- If $p<1$, $\varepsilon^{1-p}\to0$ and the integral converges to $\frac1{1-p}$.
- If $p>1$, $\varepsilon^{1-p}\to+\infty$: divergence.
- If $p=1$: $\int_\varepsilon^1\frac{dx}x=-\ln\varepsilon\to+\infty$: divergence.

So $\int_0^1x^{-p}dx$ converges iff $p<1$.

Remark. These two families are the benchmark for all comparison tests. Notice that no $p$ makes both converge: $\int_0^{+\infty}x^{-p}dx$ always diverges.