We know from definition,

$F(s) = L[f(t)] = \int\limits_{0}^{\infty}e^{-st}f(t)dt$

Here within case s = 0, $f(t) = \frac{cos6t - cos4t}{t}$

We first take, $L[cos6t - cos4t] = \frac{s}{s^2 + 6^2} - \frac{s}{s^2 + 4^2}$

$L[cos6t - cos4t] = \frac{s}{s^2 + 36} - \frac{s}{s^2 + 4^2}$

Now by using division 't' property,

$\therefore L\bigg[\frac{cos6t - cos4t}{t}\bigg] = \int\limits_{s}^{\infty}\frac{s}{s^2 + 36} - \frac{s}{s^2 + 16}ds$

$\therefore L\bigg[\frac{cos6t - cos4t}{t}\bigg] = \bigg[\frac{1}{2}log(s^2 + 36) - \frac{1}{2}log(s^2 + 16)\bigg]_{s}^{\infty}$

$\therefore L\bigg[\frac{cos6t - cos4t}{t}\bigg] = -\frac{1}{2}log(s^2 + 36) + \frac{1}{2}log(s^2 + 16)$

$\therefore L\bigg[\frac{cos6t - cos4t}{t}\bigg] = \frac{1}{2}log\bigg(\frac{s^2 + 16}{s^2 + 36}\bigg)$

Now substituting s = 0

$\therefore \int\limits_{0}^{\infty}\frac{cos6t - cos4t}{t}dt = \frac{1}{2}log\bigg(\frac{16}{36}\bigg)$

$\therefore \int\limits_{0}^{\infty}\frac{cos6t - cos4t}{t}dt = log\sqrt{\frac{16}{36}} = log\frac{4}{6}$

$\therefore \int\limits_{0}^{\infty}\frac{cos6t - cos4t}{t}dt = log\frac{2}{3}$