complex analysis - Integration using residues

For the following problem from Brown and Churchill's Complex Variables, 8ed., section 84

Show that

$$\int_0^\infty\frac{\cos(ax) - \cos(bx)}{x^2} \mathrm{d}x= \frac{\pi}{2}(b-a)$$

where $a$ and $b$ are positive, non-zero constants, by integrating about a suitable indented contour. The contour in question is the upper half of an annulus bisected by the $x$-axis with an outer radius of $R$ and in inner radius of $\delta$, such that the singularity at $x = 0$ is completely avoided by the integration.

I would expect the evaluation to involve the Cauchy-Goursat theorem to show that the integral about the entire region $\mathscr{D}$ which is enclosed by the half-annulus is $0$.

However I'm having difficulty constructing the appropriate complex analogue.

For example I know that the function $f(x) = \frac{1}{x^2}$ has the complex analogue $f(z) = \frac{1}{z^2}$ and that $\cos(x)$ can be obtained by extracting the real portion of the exponential function i.e. $\cos(x) = \mathrm{Re}(e^{ix})$.

What is the appropriate analogue for this function? Can you show that it would work?