Let $a,b>0$, without loss of generality we assume $a. Let $0<\varepsilon. First we split up the integral:
$\int_{\varepsilon}^R \frac{f(ax)-f(bx)}{x} \, dx = \int_{\varepsilon}^R \frac{f(ax)}{x} \, dx- \int_{\varepsilon}^R \frac{f(bx)}{x} \, dx$ where $ \int_{\varepsilon}^R \frac{f(ax)}{x} \, dx \stackrel{z:=a \cdot x}{=} \int_{a \cdot \varepsilon}^{a \cdot R} \frac{f(z)}{\frac{z}{a}} \cdot \frac{1}{a} \, dz = \int_{a \cdot \varepsilon}^{a \cdot R} \frac{f(z)}{z} \, dz$ (similarily for the second integral), thus
$\int_{\varepsilon}^R \frac{f(ax)-f(bx)}{x} \, dx = \underbrace{\int_{a \varepsilon}^{b \varepsilon} \frac{f(z)}{z} \, dz}_{=:I_1} - \underbrace{\int_{a \cdot R}^{b \cdot R} \frac{f(z)}{z} \, dz}_{=:I_2}$
- We have $I_1 = \int_{a \varepsilon}^{b \varepsilon} \frac{f(z)}{z} \, dz \stackrel{y:= \frac{z}{\varepsilon}}{=} \int_a^b \frac{f(\varepsilon \cdot y)}{y} \, dy$ Since $a,b>0$ (thus $[a,b] \ni y \mapsto \frac{1}{y} \in L^1([a,b])$) and $f(\varepsilon \cdot y) \to L$ as $\varepsilon \to 0$ for all $y \in [a,b]$ we can apply dominated convergence and obtain $I_1 \to \int_a^b \frac{L}{y} \, dy = L \cdot (\log b-\log a) \qquad (\varepsilon \to 0)$
- We want to prove $I_2 \to 0$ as $R \to \infty$. Let $\delta>0$. Define $I_R := \int_a^R \frac{f(z)}{z} \, dz$ Since $\int_a^{\infty} \frac{f(z)}{z} \, dz$ converges by assumption, we know that $I_R$ is a cauchy-sequence, i.e. there exists $S_0$ such that for all $S,T \geq S_0$: $|I_S-I_T| \leq \delta \tag{1}$ Now choose $R_0>0$ such that $a \cdot R_0 \geq S_0$. Then we obtain from (1) for all $R \geq R_0$: $|I_2| = |I_{b \cdot R}-I_{a \cdot R}| \leq \delta$ i.e. $I_2 \to 0$ as $R \to \infty$.
Adding all up we obtain
$\int_0^\infty \frac{f(ax)-f(bx)}{x} \, dx = L \cdot (\log b- \log a) = f(0) \cdot (\log b-\log a)$