We can write your integral as
$I=\int_0^\infty \int_{\alpha x}^\infty \int_{\beta x}^\infty \frac{\cos u \cos v}{uv } du\: dv\: dx,$
or, what is the same,
$\int_0^\infty \int_{x}^\infty \int_{x}^\infty \frac{\cos \beta u \cos \alpha v}{uv } du\: dv\: dx.$
Now the region of integration is $\{(x,u,v): x
which, up to a set of measure zero, we can write as the disjoint union of the regions $\{(x,u,v): x and $\{(x,u,v): x
Now for example, for the second region we have
$\int_0^\infty \int_{x}^\infty \int_{v}^\infty \square\: du\: dv\: dx = \int_0^\infty \int_{0}^u \int_{0}^v \square\: dx\: dv\: du $
and for us, this gives
$ \int_0^\infty \int_{0}^u \int_{0}^v \frac{\cos \beta u \cos \alpha v}{uv } dx\: dv\: du = \int_0^\infty \int_{0}^u \frac{\cos \beta u \cos \alpha v}{u } dv\: du = \int_0^\infty \frac{\cos \beta u \sin \alpha u}{\alpha u } du$
Now switching the roles of $u$, $v$, and the roles of $\alpha$ and $\beta$, and adding the resulting two integrals, we get that
$I(\alpha, \beta)=\int_0^\infty \frac{\beta \cos \beta t \sin \alpha t + \alpha \sin \beta t \cos \alpha t}{\alpha \beta t } dt.$
Someone may be able to take it from there. The resemblance with the sine integral suggests to me that adapting one of the methods used to evaluate $\int_0^\infty \frac{\sin t}{t} dt$ may work.
Edit: thanks to Didier Piau, here is the final step of the solution (direct quote from his post)
Starting from the penultimate expression in Bruno's solution, namely $I(\alpha, \beta)=\int_0^\infty \frac{\beta \cos \beta t \sin \alpha t + \alpha \sin \beta t \cos \alpha t}{\alpha \beta t } \mathrm dt,$ let us use the trigonometric relations $ 2\cos \beta t \sin \alpha t =\sin(\alpha+\beta)t+\sin(\alpha-\beta)t,\quad 2\sin \beta t \cos \alpha t =\sin(\alpha+\beta)t+\sin(\beta-\alpha)t, $ and the fact that for every $\gamma\ne0$, the change of variables $t\to\gamma t$ yields $ \int_0^\infty \frac{\sin \gamma t}{t} \mathrm dt=\text{sgn}(\gamma)\int_0^\infty \frac{\sin t}{t} \mathrm dt=\text{sgn}(\gamma)\frac{\pi}2, $ where $\text{sgn}(\gamma)$ is $+1$ if $\gamma\gt0$, $-1$ if $\gamma\lt0$, and $0$ if $\gamma=0$. This yields $ I(\alpha,\beta)=\frac\pi{4\alpha}(1+\text{sgn}(\alpha-\beta))+\frac\pi{4\beta}(1+\text{sgn}(\beta-\alpha)). $ This expression is symmetric in $(\alpha,\beta)$, as it should be. If $\alpha\gt\beta$, the second term is zero and the first one is $\pi/(2\alpha)=\pi/(2\max(\alpha,\beta))$. Finally, if $\alpha=\beta$, both terms are $\pi/(4\alpha)=\pi/(4\beta)$ hence the sum is $\pi/(2\alpha)=\pi/(2\beta)$. This proves the desired formula.