We make use of the identity
$$ \sum_{n=-\infty}^{\infty} \frac{1}{a^{2} - (x + n\pi)^{2}} = \frac{\cot(x+a) - \cot(x-a)}{2a}, \quad a > 0 \text{ and } x \in \Bbb{R}. $$
Then for $\alpha, \beta > 0$ it follows that
\begin{align*}
I := \mathrm{PV}\int_{0}^{\infty} \frac{\log\cos^{2}(\alpha x)}{\beta^{2} - x^{2}}
&= \frac{1}{2} \mathrm{PV} \int_{-\infty}^{\infty} \frac{\log\cos^{2}(\alpha x)}{\beta^{2} - x^{2}} \, dx \\
&= \frac{\alpha}{2} \mathrm{PV} \int_{-\infty}^{\infty} \frac{\log\cos^{2}x}{(\alpha\beta)^{2} - x^{2}} \, dx \\
&= \frac{\alpha}{2} \sum_{n=-\infty}^{\infty} \mathrm{PV} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\log\cos^{2}x}{(\alpha\beta)^{2} - (x+n\pi)^{2}} \, dx \\
&= \frac{\alpha}{2} \mathrm{PV} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \sum_{n=-\infty}^{\infty} \frac{1}{(\alpha\beta)^{2} - (x+n\pi)^{2}} \right) \log\cos^{2}x \, dx \\
&= \frac{1}{4\beta} \mathrm{PV} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cot(x+\alpha\beta) - \cot(x-\alpha\beta)) \log\cos^{2}x \, dx,
\end{align*}
where interchanging the order of integration and summation is justified by Tonelli's theorem applied to the summation over large indices $n$. Then
\begin{align*}
I
&= \frac{1}{4\beta} \mathrm{PV} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cot(x+\alpha\beta) - \cot(x-\alpha\beta)) \log\cos^{2}x \, dx \\
&= \frac{1}{2\beta} \mathrm{PV} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cot(x+\alpha\beta) - \cot(x-\alpha\beta)) \log\left|2\cos x\right| \, dx \tag{1}
\end{align*}
Here, we exploited the following identity to derive (1).
$$ \mathrm{PV} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cot(x+a) \, dx = 0 \quad \forall a \in \Bbb{R}. $$
Now with the substitution $z = e^{2ix}$ and $\omega = e^{2i\alpha\beta}$, it follows that
\begin{align*}
I
&= \frac{1}{2\beta} \Re \mathrm{PV} \int_{|z|=1} \left( \frac{\bar{\omega}}{z - \bar{\omega}} - \frac{\omega}{z - \omega} \right) \log(1 + z) \, \frac{dz}{z}. \tag{2}
\end{align*}
Now consider the following unit circular contour $C$ with two $\epsilon$-indents $\gamma_{\omega,\epsilon}$ and $\gamma_{\bar{\omega},\epsilon}$.
Then the integrand of (2)
$$ f(z) = \left( \frac{\bar{\omega}}{z - \bar{\omega}} - \frac{\omega}{z - \omega} \right) \frac{\log(1 + z)}{z} $$
is holomorphic inside $C$ (since the only possible singularity at $z = 0$ is removable) and has only logarithmic singularity at $z = -1$. So we have
$$ \oint_{C} f(z) \, dz = 0. $$
This shows that
\begin{align*}
I
&= \frac{1}{2\beta} \Re \lim_{\epsilon \downarrow 0} \left( \int_{-\gamma_{\omega,\epsilon}} f(z) \, dz + \int_{-\gamma_{\bar{\omega},\epsilon}} f(z) \, dz \right) \\
&= \frac{1}{2\beta} \Re \left( \pi i \mathrm{Res}_{z=\omega} f(z) + \pi i \mathrm{Res}_{z=\bar{\omega}} f(z) \right) \\
&= \frac{1}{2\beta} \Re \left( - \pi i \log(1 + \omega) + \pi i \log(1 + \bar{\omega}) \right) \\
&= \frac{\pi}{\beta} \arg(1 + \omega)
= \frac{\pi}{\beta} \arctan(\tan (\alpha \beta)).
\end{align*}
In particular, if $\alpha\beta < \frac{\pi}{2}$ then we have
$$ I = \pi \alpha. $$
But due to the periodicity of $\arg$ function, this function draws a scaled saw-tooth function for $\alpha > 0$. Of course, $I$ is an even function of both $\alpha$ and $\beta$, so the final result is obtained by even extension of this saw-tooth function.
Forgive me for my previous somewhat vague answer.
This is not so much an answer but rather a suggestion for a path to the answer,
First off, I believe there are places where we have to be a bit careful. I will mark them with an $(\ast)$ below:
\begin{align*}
\int_0^\infty\int_\alpha^\beta \sin(xy) dydx &= \int_\alpha^\beta\int_0^\infty \tag{$\ast_1$} \sin(xy)dxdy \\
&=\int_\alpha^\beta \frac{-\cos(xy)}{y} \Bigr\rvert_0^\infty dy \\
&=-\int_\alpha^\beta \lim_{a\to \infty}\left(\frac{\cos(ay)}{y}\right) - \lim_{a\to 0}\left(\frac{\cos(ay)}{y}\right) dy \\
&= \int_\alpha^\beta \lim_{a\to 0}\left(\frac{\cos(ay)}{y}\right) dy -
\int_\alpha^\beta \lim_{a\to \infty}\left(\frac{\cos(ay)}{y}\right) dy \\
&= \int_\alpha^\beta \frac{1}{y}dy - \int_\alpha^\beta \lim_{a\to \infty}\left(\frac{\cos(ay)}{y}\right) dy \\
&= \ln(\beta/\alpha) - \int_\alpha^\beta \lim_{a\to \infty}\left(\frac{\cos(ay)}{y}\right) dy \\
&= \ln(\beta/\alpha) - \lim_{a\to \infty}\int_\alpha^\beta \left(\frac{\cos(ay)}{y}\right) dy \tag{$\ast_2$} \\
&= \ln(\beta/\alpha)
\end{align*}
Now, before I address the two asteriks I will explain how we got from the second to last line to the last line. Inspired by Conrad's comment, The Riemman-Lebesgue Lemma states the following (I am citing this link (http://mathworld.wolfram.com/Riemann-LebesgueLemma.html) as my source for the statement of this lemma)
If $f(x)$ is integrable on $[-\pi, \pi]$, then
\begin{equation*}
\lim_{t\to\infty}\int_{-\pi}^\pi f(x)\sin(tx)dx = 0
\end{equation*}
In our case $f(x) = 1/y$. However I am unsure if we can generalize this lemma to any interval $[\alpha,\beta]$.
Now, as for the asteriks, for the first one it is not trivial that we can apply Fubini's theorem here and I THINK that we have to apply some measure theory to make this change of order of integration a little bit more rigorous. The same goes for the second asteriks. It is not trivial that we can move the limit outside of the integral and I think we have to apply another result from measure theory to make this second asteriks more rigorous. This thread (Can a limit of an integral be moved inside the integral?) discusses which results might be useful.
Best Answer
\begin{align} \int_{0}^{\infty}e^{-tx}\frac{\cos (\alpha x)-\cos (\beta x)}{x}dx &= \int_{0}^{\infty}dx\int_{\alpha}^{\beta}e^{-tx}\sin (yx)dy\\ &=\int_{\alpha}^{\beta}dy\int_{0}^{\infty}e^{-tx}\sin(yx)dx\\ &=\int_{\alpha}^{\beta}dy\dfrac{y}{t^2+y^2}\\ &=\dfrac12\ln\dfrac{t^2+\beta^2}{t^2+\alpha^2} \end{align} now let $t=0$.