calculus - How to show that $intlimits_1^{infty} frac{1}{x}dx$ diverges(without using the harmonic series)?

I was reading up on the harmonic series,
$H=\sum\limits_{n=1}^\infty \frac{1}{n}$, on Wikipedia, and it's divergent, as can be shown by a comparison test using the fact that

$H=1+\frac{1}{2}+(\frac{1}{3}+\frac{1}{4})+(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})+...\geq 1+\frac{1}{2}+(\frac{1}{4}+\frac{1}{4})+(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8})+...=1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+...,$ where the expression on the right clearly diverges.

But after this proof idea was given, the proof idea using the integral test was given. I understand why $H_n=\sum_{k=1}^n \frac{1}{k}\geq \int_1^n \frac{dx}{x}$, but how is it shown that $\int_1^\infty \frac{dx}{x}$ is divergent without using the harmonic series in the following way:
$H_n-1\leq \int_1^n \frac{dx}{x} \leq H_n$, and then using this in the following way, by comparison test:

$\lim_{n\rightarrow\infty}H_n=\infty\Rightarrow\lim_{n\rightarrow\infty}(H_n-1)=\infty\Rightarrow\lim_{n\rightarrow\infty}\int_1^n \frac{dx}{x}=\infty$.

So to summarize, is there a way to prove that $\int_1^\infty \frac{dx}{x}$ without using the fact that $H$ diverges?

Answer

Let $x = y/2.$ Then

$$\int_1^\infty\frac{dx}{x} = \int_2^\infty\frac{dy}{y}.$$

That is a contradiction unless both integrals equal $\infty.$