Integrate $\int_0^\infty\frac{\cos(\alpha x) - \cos(\beta x)}{x}dx$ This is problem 2 in UChicago's 2019 Math GRE prep worksheet: (link here)
Suppose $\alpha, \beta > 0$. Compute $$\int_0^\infty\dfrac{\cos(\alpha x) - \cos(\beta x)}{x}dx.$$
In the answer sheet (pg. 10) it states "rewrite integrand as integral of $\sin(xy)$, exchange order of integration." What does this mean exactly? What is $y$?
When I attempted this problem, I thought it made sense expanding the integrand out in Taylor series. Since
$$\cos(\alpha x) - \cos(\beta x) = - (\alpha x)^2/2! + \cdots + (\beta x)^2/2! -\cdots$$
and
$$\dfrac{\cos(\alpha x) - \cos(\beta x)}{x} = -\alpha^2x/2! + \cdots +\beta^2x/2!-\cdots.$$
But integrating this produces a messy series, so I don't think it's a viable solution for a GRE problem.
 A: Here is a quick, not-all-that-rigorous way to do it using the hint (though I think I'm giving way too much credit to the hint here). As it suggests, rewrite as $$ \int_0^\infty dx\int_\alpha^\beta dy\sin(xy).$$  Now, write this as $$ \frac{1}{i}\Im\int_0^\infty dx\int_\alpha^\beta dy e^{ixy}.$$ Now, shift the contour to give $y$ a small positive imaginary part, and change the order of integration, do the $x$ integral, and take the limit as the imaginary part goes to zero to get $$ \Im\int_\alpha^\beta \frac{1}{y}dy =\log(\beta/\alpha).$$
This is a case of the Frullani integral, but slightly more complicated due to the extra convergence issues.
A: I presume what they're trying to say is the following: We introduce $y$ as another variable in order to write this as a double integral. Then we can write
$$\int_0^\infty \frac{\cos(\alpha x) - \cos(\beta x)}{x}dx = \int_0^\infty\int_\alpha^\beta \sin(xy)\ dy\ dx $$
Then we can interchange the bounds of integration to get
$$=\int_\alpha^\beta \int_0^\infty \sin(xy)\ dx\ dy$$
But this is an incorrect solution, as we definitely can't integrate $\sin(kx)$ for any nonzero $k$ from $0$ to infinity.
We can, however, remedy this by considering limits of proper integrals as in this answer.
A: AS noticed by you this integral is convergent and it equals $$\log(b/a)$$ as per very interesting Theorems in
Proof of Frullani's theorem
