I am currently writing a short note about the proof techniques. I found a random theorem and wanted to write a proof by contradiction as an example. The theorem says
The integral $\int_{1}^{\infty}1/x^{p}\,\mathrm{d}x$ converges if and only if $p>1$.
This means that I have to prove that
$\impliedby:$ If $p>1$, then $\int_{1}^{\infty}1/x^{p}\,\mathrm{d}x$ converges, and
$\implies$: If $p\leq 1$, then $\int_{1}^{\infty}1/x^{p}\,\mathrm{d}x$ diverges.
Proving the first case is easy. The second case is contraposed implication, which is also easy to prove --- but I want to prove it by contradiction with this implication. Should I say,
(i) Assume that there exists a $p>1$ such that $\int_{1}^{\infty}1/x^{p}\,\mathrm{d}x$ diverges, or
(ii) Assume that there exists a $p\leq 1$ such that $\int_{1}^{\infty}1/x^{p}\,\mathrm{d}x$ converges?
If (ii) is correct, then the proof would be too superfluous. Is there a better way to prove it by contradiction (with or without this implication)?