I tried to compute $\int_{0}^{\infty}\frac{\cos(ax)-\cos(bx)}{x^2}dx \\ $.
I took $f(z) = \frac{\exp(aiz)-\exp(biz)}{z^2}$, since the real part of this is equal to $\frac{\cos(ax)-\cos(bx)}{x^2}.$
Then I applied contour integration. The contour $C =\gamma_{\epsilon}\cup \Gamma_{R}\cup[-R,-\epsilon)\cup(\epsilon,R] $, where
$\bullet \ \gamma_{\epsilon} $ is the semicircle $\gamma_{\epsilon} = \{z:|z| = \epsilon, \text{Im z} \geq 0 \} $
$\bullet \ \Gamma_{R} $ is the semicircle $\Gamma_{R} = \{z:|z| = R, \text{Im z} \geq 0 \}$
Then $\text{lim}_{\epsilon \to 0 } \int_{\gamma_{\epsilon}}\frac{\exp(aiz)-\exp(biz)}{z^2} = \pi(a-b)\\$
Because I do it in the way you normally calculate line integrals:
$\text{lim}_{\epsilon \to 0 } \int_{\gamma_{\epsilon}}\frac{\exp(aiz)-\exp(biz)}{z^2} \\ $
$\text{lim}_{\epsilon \to 0 } \int_{0}^{\pi}\frac{\exp(ai(-\epsilon\cdot \exp(it)))-\exp(bi(-\epsilon\cdot \exp(it)))}{(-\epsilon\cdot \exp(it))^2} \cdot |-i \cdot \epsilon \cdot \exp(it)| dt\\ $
$\text{But I don't know how to compute this}.\\ \text{What is the best way to show that the limit} \ \epsilon \to 0 \ \text{of the integral over} \ \gamma_\epsilon \ \text{ goes to} \ \pi(a-b)\text{?}$
$$\frac{e^{aiz}-e^{biz}}{z^2}=\frac{(a-b)i}{z}+f(z),\quad f(z):=\frac{(e^{aiz}-1-aiz)-(e^{biz}-1-biz)}{z^2}$$ and $f(z)$ has a removable singularity at $z=0$, hence $\lim\limits_{\epsilon\to 0}\int_{\gamma_\epsilon}f(z)\,dz=0$ and $$\lim_{\epsilon\to 0}\int_{\gamma_\epsilon}\frac{e^{aiz}-e^{biz}}{z^2}\,dz=(a-b)i\color{gray}{\lim_{\epsilon\to 0}}\int_{\gamma_\epsilon}\frac{dz}{z}=\pi(\color{red}{b-a}).$$