We know from definition,

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

Here withing 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] = $\frac{s}{s^2 + 6^2} - \frac{s}{s^2 + 4^2}$

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

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

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

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