$\displaystyle Consider\\[2ex] L[\cos{at-\cos{bt}}]=L[\cos{at]-L[\cos{bt]}} \\[2ex] \displaystyle =\ \frac{s}{s^2+a^2}-\frac{s}{s^2+b^2} \\[4ex] \displaystyle \therefore{}\ L[\cos{at-\cos{bt}}]\\[2ex] \displaystyle = \int_0^{\infty{}}[\frac{s}{s^2+a^2}-\frac{s}{s^2+b^2}]ds \\[2ex] \displaystyle =\frac{1}{2}\int_0^{\infty{}}\left[\frac{s}{s^2+a^2}-\frac{s}{s^2+b^2}\right]ds \\[2ex] \displaystyle =\frac{1}{2}{\left[\log{\left(s^2+a^2\right)}-\log⁡(s^2+b^2)\right]}_s^{\infty{}} \\[2ex] \displaystyle =\frac{1}{2}{\left[\log{\left(\frac{s^2+a^2}{s^2+b^2}\right)}\right]}_s^{\infty{}} \\[2ex] \displaystyle =\frac{1}{2}\left[\ 0-\log{\left(\frac{s^2+a^2}{s^2+b^2}\right)}\right] \\[2ex] $

$\displaystyle \therefore{}\int_0^{\infty{}}e^{-st}\left(\frac{\cos{at-\cos{bt}}}{t}\right)dt\\[3ex] \displaystyle =\frac{1}{2}\log{\left(\frac{s^2+b^2}{s^2+a^2}\right)} \\[2ex] $

$\displaystyle Put\ s=0, \\[2ex] \displaystyle \therefore{}\int_0^{\infty{}}\frac{\cos{at-\cos{bt}}}{t}dt\\[2ex] \displaystyle =\frac{1}{2}\log\left(\frac{0+b^2}{0+a^2}\right) \\[2ex] \displaystyle ={\log\left(\frac{b^2}{a^2}\right)}^{\frac{1}{2}} \\[3ex] \displaystyle =\log\left(\frac{b}{a}\right)$