Show that $$\int\limits_0^\infty\frac{1}{t}(\cos(at)-\cos(bt))dt=\ln(b/a),\,a,b>0.$$
Thanks to wikipedia I know that
$$\int\limits_0^\infty\frac{1}{t}(\cos(at)-\cos(bt))\,dt \overset{?}{=}\int\limits_{0}^\infty\left[\frac{p}{p^2+a^2}-\frac{p}{p^2+b^2}\right]=\left.\frac{1}{2}\ln\left[\frac{p^2+a^2}{p^2+b^2}\right]\right|_{0}^\infty=\ln b-\ln a.$$ I am having a hard time understanding the jump between the equality with the question mark above it.
Write
$$\frac1{t} = \int_0^{\infty} dp \, e^{-p t}$$
Then the integral is
$$\begin{align}\int_0^{\infty} dt \, \frac{\cos{a t}-\cos{b t}}{t} &= \int_0^{\infty} dt \, (\cos{a t} - \cos{b t})\int_0^{\infty} dp \, e^{-p t} \\ &= \int_0^{\infty} dp \, \left (\int_0^{\infty} dt \, \cos{a t} \, e^{- p t} - \int_0^{\infty} dt \, \cos{b t} \, e^{- p t}\right ) \\ &= \int_0^{\infty} dp \, \left (\frac{p}{p^2+a^2} - \frac{p}{p^2+b^2} \right ) \end{align} $$
It seems you got the rest. Note that, in the second line, we can reverse the order of integration because the integrals involved are convergent. Also note that, in the third line, as @DrMV points out, the expressions are Laplace transforms of the respective cosine functions.
Have you heard about Frullani's Integrals? Otherwise you can see here for different techniques.
