real analysis - Show $f(x) = 1/x$ is in $L^2left([1, +infty)right)$ but not in $L^1left([1, +infty)right)$.

Proposition

$f(x) = 1/x$ is in $L^2\left([1, +\infty)\right)$ but not in $L^1\left([1, +\infty)\right)$.

Discussion

So my issue here is that I don't know how to use infinity in Lebesgue integration.

It is intuitive (I think) that evaluation of the improper Riemann integrals

\begin{align}
\int_1^\infty \left|f(x)\right| &= \int_1^\infty \frac{1}{x} = \lim_{c \to \infty} \ln c = + \infty \\ \\
\int_1^\infty \left|f(x)\right|^2 &= \int_1^\infty \frac{1}{x^2} = 1 - \lim_{c \to \infty} \frac{1}{c} = 1
\end{align}

would imply our proposition, but I've only seen $L^p$-spaces defined in the sense of Lebesgue integrals. So when I get to these steps:

\begin{align}
\int_{[1, \infty)} \left|f(x)\right| &= \int_{[1, \infty)} \frac{1}{x} = \cdots \\ \\
\int_{[1, \infty)} \left|f(x)\right|^2 &= \int_{[1, \infty)} \frac{1}{x^2} = \cdots
\end{align}

I'm not sure how to proceed. I'm guessing we need an argument for switching between the two types of integration, which I've read up on a little bit, but am not sure how to apply here in the improper case.