The integral is zero since

$$\int_0^\infty \frac{\sin(ax)}{x}\,dx=\frac{\pi}{2}\text{sgn}(a)$$

Thus, we have

$$\int_0^\infty \frac{\sin(ax)-\sin(bx)}{x}\,dx=\frac{\pi}{2}(\text{sgn}(a)-\text{sgn}(b))$$

which is zero whenever $a$ and $b$ are of the same sign.

Now, this doesn't answer the question "Why doesn't the formula for Frullani's Integral seem to apply here?" Certainly, $\lim_{L\to \infty} \sin(Lx)$ fails to exist and so $f(\infty)$ is meaningless.

This compels one to go back in the development of Frullani to review the origin of the term $f(\infty)$. In the development of the formula, we start with the integral

$$\begin{align}
\int_\epsilon^L \int_a^b f'(xy)\,dy\,dx&=\int_\epsilon^L \frac{f(bx)-f(ax)}{x}\,dx\\\\
&=\int_a^b \frac{f(Ly)-f(\epsilon y)}{y}\,dy
\end{align}$$

Letting $L\to \infty$ and $\epsilon \to 0$ we find that if the integral $\int_0^\infty \frac{f(bx)-f(ax)}{x}\,dx$ exists, then

$$\int_0^\infty \frac{f(bx)-f(ax)}{x}\,dx=\lim_{L\to \infty}\lim_{\epsilon\to 0}\int_a^b \frac{f(Lx)-f(\epsilon x)}{x}\,dx$$

We are tempted to interchange the order of the limits and integral. However, if either $\lim_{x\to 0}f(x)$ or $\lim_{x\to \infty}f(x)$ fails to exist, then that interchange is obviously invalid.

However, since $\frac1x$ is integrable on $[a,b]$, $0<a<b$, then the Riemann-Lebesgue Lemma guarantees that for $f(x)=\sin(x)$

$$\begin{align}
\lim_{L\to \infty}\int_a^b \frac{f(Lx)}{x}\,dx&=\lim_{L\to \infty}\int_a^b \frac{\sin(Lx)}{x}\,dx\\\\
&=0
\end{align}$$

Therefore, we find that

$$\int_0^\infty \frac{\sin(bx)-\sin(ax)}{x}\,dx=-\lim_{\epsilon \to 0}\int_a^b \frac{\sin(\epsilon x)}{x}\,dx=0$$

as expexted!

**NOTE:**

In THIS ANSWER, I developed a generalization of Frullani's Integral applicable for complex values of $a$ and $b$.