I am not sure if the following way is valid to evaluate
$$\int_{0}^{\infty}\frac{\cos(\alpha x)-\cos(\beta x)}{x}\text{d}x$$
But I am asking if the following is valid or not?
Edit:
Note that $\int_{0}^{\infty}\frac{f(t)}{t}\text{d}t=\int_{0}^{\infty}\mathcal{L}\text{\{}f(t)\text{\}}\text{d}s$
$\int_{0}^{\infty}\frac{\cos(\alpha x)}{x}\text{d}x=\int_{0}^{\infty}\mathcal{L}\text{\{}\cos(\alpha x)\text{\}}\text{d}s=\int_{0}^{\infty}\frac{s}{s^2+\alpha^2}\text{d}s=\frac{1}{2}\log(s^2+\alpha^2)$, and
$\int_{0}^{\infty}\frac{\cos(\beta x)}{x}\text{d}x=\int_{0}^{\infty}\mathcal{L}\text{\{}\cos(\beta x)\text{\}}\text{d}s=\int_{0}^{\infty}\frac{s}{s^2+\beta^2}\text{d}s=\frac{1}{2}\log(s^2+\beta^2)$
Taking the difference, we get:
$\frac{1}{2}\log(\frac{s^2+\alpha^2}{s^2+\beta^2})$
Now applying the limits $s=0$ to $s \rightarrow \infty$, we get:
$\frac{1}{2}\log(1)-\frac{1}{2}\log(\frac{\alpha^2}{\beta^2})=\log(\frac{\beta}{\alpha})$
Having the right answer does not mean that the way is right.
Your help would be appreciated. Thanks!