Suppose $\alpha, \beta>0$, I want to compute \begin{align*} \int_{0}^{\infty}\frac {\cos(\alpha x)-\cos(\beta x)}{x}\, dx. \end{align*} I've change it to the following \begin{align*} \int_{0}^{\infty}\int_{\alpha}^{\beta}\sin (xy)\, dy\, dx. \end{align*} and get: \begin{align*} \ln{\frac {\beta}{\alpha}}-\lim_{a\to \infty} \int_{\alpha}^{\beta}\frac {\cos{ay}}{y}\, dy. \end{align*}
The final answer is $\ln{\frac {\beta}{\alpha}}$. I wonder how to argue the limit part is $0$. Thanks.
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.
