There is a claim that is slightly more general.
Let $f$ be such that $\int_a^b f$ exists for each $a,b>0$. Suppose that $$A=\lim_{x\to 0^+}x\int_{x}^1 \frac{f(t)}{t^2}dt\\B=\lim_{x\to+\infty}\frac 1 x\int_1^x f(t)dt$$ exist.
Then $$\int_0^\infty\frac{f(ax)-f(bx)}xdx=(B-A)\log \frac ab$$
PROOF Define $xg(x)=\displaystyle \int_1^x f(t)dt$. Since $g'(x)+\dfrac{g(x)}x=\dfrac{f(x)}x$ we have $$\int_a^b \frac{f(x)}xdx=g(b)-g(a)+\int_a^b\frac{g(x)}xdx$$
Thus for $T>0$
$$\int_{Ta}^{Tb} \frac{f(x)}xdx=g(Tb)-g(Ta)+\int_{Ta}^{Tb}\frac{g(x)}xdx$$
But
$$\int_{Ta}^{Tb}\frac{g(x)}xdx-B\int_a^b \frac{dx}x=\int_a^b\frac{g(Tx)-B}xdx$$
Thus $$\lim_{T\to+\infty}\int_{Ta}^{Tb}\frac{g(x)}xdx=B\log\frac ba$$ so
$$\lim_{T\to+\infty}\int_{Ta}^{Tb}\frac{f(x)}xdx=B\log\frac ba$$
It follows, since $$\int_1^T\frac{f(ax)-f(bx)}xdx=\int_{bT}^{aT}\frac{f(x)}xdx+\int_a^b \frac{f(x)}xdx$$ (note $a,b$ are swapped) that $$\int_1^\infty \frac{f(ax)-f(bx)}xdx=B\log\frac ab+\int_a^b \frac{f(x)}xdx$$
Let $\varepsilon >0$, $\hat f(x)=f(1/x)$. Then $$\int\limits_\varepsilon ^1 {\frac{{f\left( x \right)}}{x}dx} = \int\limits_1^{{\varepsilon ^{ - 1}}} {\frac{{\hat f\left( x \right)}}{x}dx} $$ and $$x\int\limits_x^1 {\frac{{f\left( t \right)}}{{{t^2}}}dt} = \frac{1}{{{x^{ - 1}}}}\int\limits_1^{{x^{ - 1}}} {\hat f\left( t \right)dt} = g\left( {{x^{ - 1}}} \right)$$
So $\hat f(t)$ is in the hypothesis of the preceding work. It follows that $$\lim_{T\to+\infty}\int\limits_1^T {\frac{{\hat f\left( {x{a^{ - 1}}} \right) - \hat f\left( {x{b^{ - 1}}} \right)}}{x}} dx = A\log \frac ba + \int\limits_{{a^{ - 1}}}^{{b^{ - 1}}} {\frac{{\hat f\left( x \right)}}{x}dx} $$
and by a change of variables $x\mapsto x^{-1}$ we get $$\int\limits_0^1 {\frac{{f\left( {ax} \right) - f\left( {bx} \right)}}{x}} dx = A\log \frac ba - \int\limits_a^b {\frac{{f\left( x \right)}}{x}dx} $$ and summing gives the desired $$\int\limits_0^\infty {\frac{{f\left( {ax} \right) - f\left( {bx} \right)}}{x}} dx = \left( {B - A} \right)\log \frac ab$$
This is due to T.M. Apostol.
OBS By L'Hôpital, if the limits at $x=0^+$ and $x=+\infty$ exist, they equal $A$ and $B$ respectively.