The following is just a speeded-up version of the answer by Davide Giraudo.
Let $f(x)$ be a real-valued function defined for $x\geq 0$.
Suppose that $f(x)$ is Riemann integrable on every bounded interval
of nonnegative real numbers,
that $f(x)$ is continuous at $x=0$,
and that the limit $f(\infty):=\lim_{x\to\infty} f(x)$ exists
(as a finite quantity).
If $a>0$ and $b>0$, then the integral
\begin{equation*}
\int_{\,0}^{\,\infty} \frac{f(ax)-f(bx)}{x} dx \tag{1}
\end{equation*}
exists and is equal to $\bigl(f(\infty)-f(0)\bigr)\,\ln(a/b)$.
The assertion that the integral $(1)$ exists means
that the following limit exists,
\begin{equation*}
\lim_{l\to0,\,h\to\infty} \int_{\,l}^{\,h} \frac{f(ax)-f(bx)}{x} dx
\end{equation*}
where $l$ approaches $0$ independently of $h$ approaching $\infty$,
and that this limit is the integral $(1)$.
Proof.$\,$ Assume that $a\geq b\,$; this does not lose us any generality.
Let $0<l<h\,$.
The change of variables $ax=by$ shows that
\begin{equation*}
\int_{\,l/a}^{\,h/a}\frac{f(ax)}{x}dx
~=~ \int_{\,l/b}^{\,h/b}\frac{f(by)}{y}dy~,
\end{equation*}
so we have
\begin{align*}
\int_{\,l/a}^{\,h/a}\frac{f(ax)-f(bx)}{x}dx
&~=~ \int_{\,l/b}^{\,h/b}\frac{f(bx)}{x}dx
~-~ \int_{\,l/a}^{\,h/a}\frac{f(bx)}{x}dx \\
&~=~ \int_{\,h/a}^{\,h/b}\frac{f(bx)}{x}dx
~-~ \int_{\,l/a}^{\,l/b}\frac{f(bx)}{x}dx~;
\end{align*}
we write the difference in the second line as $I(h) - I(l)$.
Let $\varepsilon>0$.
There exists $l_\varepsilon>0$ such that
$f(0)-\varepsilon\leq f(bx)\leq f(0)+\varepsilon$
for $0\leq x\leq l_\varepsilon/b$.
Since by assumption $l/a\leq l/b$ for every $l>0$, we obtain the estimate
\begin{equation*}
\int_{\,l/a}^{\,l/b}\frac{f(0)-\varepsilon}{x}dx
~\leq~ I(l)
~\leq~ \int_{\,l/a}^{\,l/b}\frac{f(0)+\varepsilon}{x}dx
\qquad\qquad \text{for $0<l\leq l_\varepsilon$}~,
\end{equation*}
that is,
\begin{equation*}
\bigl(f(0)-\varepsilon\bigr)\,\ln\frac{a}{b}
~\leq~ I(l)
~\leq~ \bigl(f(0)+\varepsilon\bigr)\,\ln\frac{a}{b}
\qquad\qquad \text{for $0<l\leq l_\varepsilon$}~.
\end{equation*}
In other words, $I(l)$ converges to $f(0)\,\ln(a/b)$ as $l$ approaches $0$.
In the same way we see that $I(h)$ converges to $f(\infty)$
as $h$ approaches $\infty$.$\,$ Done.