How do I calculate the integral $\int^{\infty}_{0}\frac{\cos(ax)-\cos(bx)}{x}dx$ without residues? Consider the integral:

$$\int^{\infty}_{0}\frac{\cos(ax)-\cos(bx)}{x}\,dx$$

I found this problem in a subject test problem set and I'm unsure how to solve it without using the residue theorem. 

I've simplified it to $$\int^{\infty}_{0}-2\frac{\sin\left(\frac{(a+b)x}{2}\right)\sin\left(\frac{(a-b)x}{2}\right)}{x}\,dx$$ but still unsure where to go from here. 

There's a hint in the problem that says: " Rewrite numerator as integral over sine, estimate, and exchange order of integration." but I don't see a helpful bound to use. Bounding using the fact that $|\sin|\le 1$ hasn't gotten me anywhere helpful. 

Any ideas?
Notice that:
$$\frac{\cos (ax)-\cos (bx)}{x}=\int_{a}^{b}{\sin \left( xs \right)ds}$$
If you define
$$I\left( t \right)=\int_{0}^{\infty }{{{e}^{-tx}}\frac{\cos (ax)-\cos (bx)}{x}}dx$$
Then
$$I\left( t \right)=\int_{0}^{\infty }{{{e}^{-tx}}\left\{ \int_{a}^{b}{\sin \left( xs \right)ds} \right\}}dx$$
Reversing the order of integration gives:
$$I\left( t \right)=\int_{a}^{b}{\left\{ \int_{0}^{\infty }{{{e}^{-tx}}\sin \left( xs \right)dx} \right\}}ds$$
integration by parts twice gives 
$$\int{{{e}^{-tx}}\sin \left( xs \right)dx=\frac{{{e}^{-tx}}\left( s\sin \left( xs \right)-x\cos \left( xs \right) \right)}{{{t}^{2}}+{{s}^{2}}}}$$
So
$$I\left( t \right)=\int_{a}^{b}{\left[ \frac{{{e}^{-tx}}\left( s\sin \left( xs \right)-x\cos \left( xs \right) \right)}{{{t}^{2}}+{{s}^{2}}} \right]}_{0}^{\infty }ds=\int_{a}^{b}{\frac{s}{{{t}^{2}}+{{s}^{2}}}ds}$$
Finally the change of variable $u={{t}^{2}}+{{s}^{2}}$ reduces the integral
$$I\left( t \right)=\frac{1}{2}\ln \left( \frac{{{t}^{2}}+{{b}^{2}}}{{{t}^{2}}+{{a}^{2}}} \right)$$
Can you guess what happens when $t\to 0$?
Split into sum of two terms, and then apply integral from table:

$$\int \frac{\cos ax}{x} dx = \ln x - \frac{(ax)^2}{2 \cdot 2!} + \frac{(ax)^4}{4 \cdot 4!}  - \frac{(ax)^6}{6 \cdot 6!} + ... 
 $$