Scale-invariance of $\int_0^\infty \frac{f(x)}{x} \ dx$ Let $f$ be some non-negative, measurable function on $[0,\infty)$. The quantity $\int_0^\infty \frac{f(x)}{x} \ dx$ is scale-invariant in the sense that, if one puts $f_c(x) := f(cx)$ for $c > 0$, then
$$\int_0^\infty \frac{f_c(x)}{x} \ dx = \int_-^\infty \frac{f(cx)}{x} \ dx = \int_0^\infty \frac{f(cx)}{cx} \ d(cx) = \int_0^\infty \frac{f(x)}{x} \ dx.$$
I am wondering if there is some compelling, e.g., physical reason for this scale-invariance. 
For example, I could think of $f$ as a mass density function for some object, say an apple if $f$ is compactly supported, located somewhere in $(0,\infty)$. Then $f_c$ is a mass density function for some scaled up version of that object. What intepretation should we give to $\int_0^\infty \frac{f(x)}{x} \ dx$ so it becomes clear that this quantity doesn't change if I double my apple, with respect to the origin?
As an indictation of what I might be looking for, consider that $\int_0^\infty f(x) x \ dx$ can be thought of as the leverage of the apple with respect to the origin (not a scale-invariant quantity). 
 A: 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.
A: In hindsight, a good explanation is probably to just note that the measure $\frac{1}{x}dx$ on $(0,\infty)$ is the pullback of the Lebesgue measure on $(-\infty,\infty)$ through the natural logarithm map $\log : (0,\infty) \to (-\infty,\infty)$. So, the scale-invariance of $\frac{1}{x}dx$ is another just manifestation of the translation-invariance of Lebesgue measure. Put differently, $\frac{1}{x}dx$ is the Haar measure for the multiplicative group $(0,\infty)$. 
