Show that a certain integral involving a continuous and compactly supported function $f$ converges to a constant times $f(0)$ Let $f:[0,\infty) \to \mathbb{R}$ be continuous with compact support. Prove that
$$
\lim_{\epsilon \to 0^+} \int_\epsilon^\infty \frac{f(ax) - f(bx)}{x} dx = f(0)\log(a/b)
$$
for all $a,b > 0$.
Given that there's a $1/x$, it seems like the trick is to some sort of $u$ sub to perhaps put $a$ and $b$ in the limits, put $\epsilon$ inside $f(x)$ and then pass the limit through by continuity.
 A: Suppose $0<a<b.$ Then
$$\int_{a}^{\infty} \frac{f(x)}{x}\,dx = \int_{a\epsilon}^{\infty} \frac{f(x)}{x}\,dx.$$
Similarly for the integral over $[b,\infty).$ Subtracting, we get
$$\tag 1 \int_{a\epsilon}^{b\epsilon} \frac{f(x)}{x}\,dx.$$ A lower bound for $(1)$ is
$$\inf_{[0,b\epsilon]}f\cdot\int_{a\epsilon}^{b\epsilon} \frac{1}{x}\,dx = \inf_{[0,b\epsilon]} f\cdot\ln(b/a).$$
Similarly, $\sup_{[0,b\epsilon]}f \cdot\ln(b/a)$ is an upper bound for $(1).$ Because $f$ is continuous at $0,$ both of these bounds $\to f(0)\ln(b/a)$ as $\epsilon\to 0^+.$ By the squeeze theorem the result follows.
A: The $\log(a/b)$ you have in your equation should be $\log(b/a)$. If $a = b$, both sides of the equation are zero, so suppose $b > a$ (the argument is similar for the case $a > b$). Let $T$ be large enough so that $aT > b\epsilon$. Then 
\begin{align}\int_{\epsilon}^T \frac{f(ax)-f(bx)}{x}\, dx &= \int_\epsilon^T \frac{f(ax)}{x}\, dx -\int_{\epsilon}^{T} \frac{f(bx)}{x}\, dx \\
&= \int_{a\epsilon}^{aT} \frac{f(x)}{x}\, dx - \int_{b\epsilon}^{bT} \frac{f(x)}{x}\, dx\\
&= \int_{a\epsilon}^{b\epsilon} \frac{f(x)}{x}\, dx - \int_{aT}^{bT} \frac{f(x)}{x}\, dx\\
&= \int_a^b \frac{f(\epsilon x)}{x}\, dx - \int_a^b \frac{f(T x)}{x}\, dx\tag{1}\label{eq1}
\end{align}
Suppose the supprt of $f$ is contained in the interval $[-d,d]$ for some $d > 0$. If $T > d/a$, then $Tx > d$ for all $x\in [a,b]$; therefore, the  function $f(Tx)/x$ is zero on $[a,b]$ whenever $T >  d$. Consequently,
$$\lim_{T\to \infty} \int_a^b \frac{f(Tx)}{x}\, dx = 0\tag{2}\label{eq2}$$
Since $f$ is continuous with compact support, it is uniformly continuous. So given $\eta > 0$, there is a $\delta > 0$ such that for all $x,y\in [0,\infty)$, $\lvert x - y\rvert < \delta$ implies $\lvert f(x)-f(y)\rvert < \eta/(b-a)$. For $0 < \epsilon < \delta/b$, we have $\lvert f(\epsilon x) - f(0)\rvert < \eta$ on $[a,b]$ and so 
$$\left\lvert \int_a^b \frac{f(\epsilon x) - f(0)}{x}\, dx\right\rvert \le \eta \quad \left(0 < \epsilon < \frac{\delta}{b}\right)$$
Therefore
$$\lim_{\epsilon\to 0^+} \int_a^b \frac{f(\epsilon x)}{x}\, dx = \int_a^b \frac{f(0)}{x}\, dx = f(0)\log\frac{b}{a}\tag{3}\label{eq3}$$
By \eqref{eq1}, \eqref{eq2}, and \eqref{eq3}, we deduce
$$\int_0^\infty \frac{f(ax)-f(bx)}{x}\, dx = f(0)\log\frac{b}{a}$$
