How to solve this improper integral? The problem is:

If $f(x)\in C[0,+\infty)$, $\displaystyle\lim_{x\to+\infty}f(x)=k\in\mathbb R$, and $b>a>0$, prove:
  $$\int_{0}^{+\infty}\frac{f(ax)-f(bx)}{x}dx=[f(0)-k]\ln(\frac ba)$$

My attempt:
This integral has two singularities, one at $x=0$ and the other at $x=+\infty$, so we'd better split it into two parts, each with only one singularity. Let $A>0$, and let $I$ denote the integral, then
$$\begin{align}
I&=\Big(\int_{0}^{A}+\int_{A}^{+\infty}\Big)\Big(\frac{f(ax)-f(bx)}{x}dx\Big)
\\&=\Big(\int_{0}^{A}+\int_{A}^{+\infty}\Big)\frac{f(ax)}{x}dx-\Big(\int_{0}^{A}+\int_{A}^{+\infty}\Big)\frac{f(bx)}{x}dx
\\&=\Big(\int_{0}^{aA}+\int_{aA}^{+\infty}\Big)\frac{f(x)}{x}dx-\Big(\int_{0}^{bA}+\int_{bA}^{+\infty}\Big)\frac{f(x)}{x}dx
\\&=\Big(\int_{0}^{aA}+\int_{bA}^{0}+\int_{aA}^{+\infty}+\int_{+\infty}^{bA}\Big)\frac{f(x)}{x}dx
\end{align}$$
If I keep going and cancel these integral limits out (and I know it's probably not doable because infinity is involved here and may play tricks on us), I simply get $I=0$,  of course that's not the desired result.
What's more, I hope to get the limit $k$ involved, but there seems no obvious way to do so.
Can you help me? Any help or hint (if not too obscure) will be appreciated. Best regards!
 A: Given $t > s > 0$,
$$\int_s^t \frac{f(ax) - f(bx)}{x}\, dx = \int_s^t \frac{f(ax)}{x}\, dx - \int_s^t \frac{f(bx)}{x}\, dx = \int_{as}^{at} \frac{f(x)}{x}\, dx - \int_{bs}^{bt} \frac{f(x)}{x}\, dx.$$
Since
\begin{align}\int_{as}^{at} \frac{f(x)}{x}\, dx - \int_{bs}^{bt} \frac{f(x)}{x}\, dx &= \int_{as}^{bs} \frac{f(x)}{x}\, dx  + \int_{bs}^{at} \frac{f(x)}{x}\, dx - \int_{bs}^{bt} \frac{f(x)}{x}\, dx \\
& = \int_{as}^{bs} \frac{f(x)}{x}\, dx - \int_{at}^{bt} \frac{f(x)}{x}\, dx\\
& = \int_a^b \frac{f(sx)}{x}\, dx - \int_a^b \frac{f(tx)}{x}\, dx\\
& = \int_a^b \frac{f(sx) - f(tx)}{x}\, dx,
\end{align}
we have
$$\int_s^t \frac{f(ax) - f(bx)}{x}\, dx = \int_a^b \frac{f(sx) - f(tx)}{x}\, dx.$$
Therefore
\begin{align}&\int_s^t \frac{f(ax) - f(bx)}{x}\, dx - [f(0) - k]\ln \frac{b}{a}\\
& =\int_a^b \frac{f(sx) - f(tx)}{x}\, dx - \int_a^b \frac{f(0) - k}{x}\, dx\\
& = \int_a^b \frac{f(sx) - f(0)}{x}\, dx - \int_a^b \frac{f(tx) - k}{x}\, dx \tag{*}.
\end{align}
Now since $f\in C[0, +\infty)$ and $\lim_{x\to +\infty} f(x)$ exists, $f$ is uniformly continuous on $[0, +\infty)$. Using uniform continuity of $f$, show that (*) tends to $0$ as $s \to 0^+$ and $t \to +\infty$.
