Scale-invariance of $\int_0^\infty \frac{f(x)}{x} \ dx$

Here's one way of thinking about it:

In a sense, this integral is a "weighted average" of a mass density function $f$, where we assign higher weight to mass in inverse proportion to its distance from $x = 0$. Let's say we have $0 < c < 1$. Then $f(cx)$ is a "squished" version of $f(x)$. On the one hand, there is only a proportion of $c$ of the original mass to spread around. On the other hand, the remaining mass is moved closer to the origin so that its importance in our weighted average increases, overall, by a factor of $1/c$.

The result is that the new weighted average is the same.

I'm not sure how to properly relate this to the leverage integral, but it seems there may be a meaningful connection to extract.

Here's a more mathematical approach:

Suppose $f$ is continuously differentiable, with $f(0) = 0$ and compact support. We then have

$$ \int_0^\infty \frac{f(x)}{x}\,dx = \ln(x)f(x) \Big|_0^\infty - \int_0^\infty f'(x)\ln(x)\,dx = -\int_0^\infty f'(x)\ln(x)\,dx $$ Now, note that $f_c' = c f'(cx)$. So, we have $$ \int_0^\infty \frac{f_c(x)}{x}\,dx = -\int_0^\infty f'(cx)\ln(x)\,(c\,dx) $$ Though I'm not sure if that's more compelling than what you already have.