$$\int_0^{\infty} \frac{\sin ax\,dx}{x(x^2+b^2)^2}=\frac{1}{b^4}\left(\int_0^{\infty}\frac{\sin ax}{x}\,dx-\int_0^{\infty}\frac{x\sin ax\,dx}{x^2+b^2}\right)-\frac{1}{b^2}\int_0^{\infty}\frac{x\sin ax\,dx}{(x^2+b^2)^2}$$
The first integral is well known. For any $a>0:$
$$\int_0^{\infty} \frac{\sin ax}{x}\,dx=\frac{\pi}{2}$$
The second, consider:
$$\begin{aligned}f(t)=\int_0^{ \infty} \frac{x\sin axt\,dx}{x^2+b^2}\,dx \Rightarrow \mathcal{L} \{ f(t)\} &=\int_0^{ \infty}e^{-st}\int_0^{ \infty}\frac{x\sin axt\,dx}{x^2+ b^2}\,dx\,dt\\&=\int_0^{ \infty}\frac{x}{x^2+ b^2}\int_0^{\infty}e^{-st}\sin axt\,dt\,dx\\&=\int_0^{\infty} \frac{ax^2}{(x^2+ b^2)(a^2x^2+s^2)} \,dx\\&= \frac{\pi}{2(s+ab)}\end{aligned}$$
$$\frac{\pi}{2}\cdot\mathcal{L}^{-1}\left\{ \frac{1}{s+ab}\right\}\Bigg|_{t=1}= \frac{\pi}{2e^{ab}}$$
The third, using the same parameter (call the function $g(t)$ now) one obtains:
$$\begin{aligned}\mathcal{L} \{ g(t)\} &=\int_0^{ \infty}\frac{x}{(x^2+ b^2)^2}\int_0^{\infty}e^{-st}\sin axt\,dt\,dx\\&=\int_0^{\infty} \frac{ax^2}{(x^2+ b^2)^2(a^2x^2+s^2)} \,dx\\&= \frac{a\pi}{4b(s+ab)^2}\end{aligned}$$
$$\frac{\pi a}{4b}\cdot\mathcal{L}^{-1}\left\{ \frac{1}{(s+ab)^2}\right\}\Bigg|_{t=1}= \frac{a\pi}{4be^{ab}}$$
Therefore:
$$\int_0^{\infty} \frac{\sin ax\,dx}{x(x^2+b^2)^2}=\frac{\pi}{2b^4}\left(1-\frac{2+ab}{2e^{ab}}\right)$$
Let $$f(y) = \int_{0}^{\infty} \frac{\sin^3{yx}}{x^3} \mathrm{d}x$$
Then,
$$f'(y) = 3\int_{0}^{\infty} \frac{\sin^2{yx}\cos{yx}}{x^2} \mathrm{d}x = \frac{3}{4}\int_{0}^{\infty} \frac{\cos{yx} - \cos{3yx}}{x^2} \mathrm{d}x$$
$$f''(y) = \frac{3}{4}\int_{0}^{\infty} \frac{-\sin{yx} + 3\sin{3yx}}{x} \mathrm{d}x$$
Therefore,
$$f''(y) = \frac{9}{4} \int_{0}^{\infty} \frac{\sin{3yx}}{x} \mathrm{d}x - \frac{3}{4} \int_{0}^{\infty} \frac{\sin{yx}}{x} \mathrm{d}x$$
Now, it is quite easy to prove that $$\int_{0}^{\infty} \frac{\sin{ax}}{x} \mathrm{d}x = \frac{\pi}{2}\mathop{\mathrm{signum}}{a}$$
Therefore,
$$f''(y) = \frac{9\pi}{8} \mathop{\mathrm{signum}}{y} - \frac{3\pi}{8} \mathop{\mathrm{signum}}{y} = \frac{3\pi}{4}\mathop{\mathrm{signum}}{y}$$
Then,
$$f'(y) = \frac{3\pi}{4} |y| + C$$
Note that, $f'(0) = 0$, therefore, $C = 0$.
$$f(y) = \frac{3\pi}{8} y^2 \mathop{\mathrm{signum}}{y} + D$$
Again, $f(0) = 0$, therefore, $D = 0$.
Hence, $$f(1) = \int_{0}^{\infty} \frac{\sin^3{x}}{x^3} = \frac{3\pi}{8}$$
Best Answer
$$\begin{aligned}\int_0^{\infty} \frac{\cos ax-\cos bx}{x^2}\,dx &=\int_0^{\infty}\int_a^{b}\frac{\sin tx}{x}\,dt\,dx \\&=\int_a^{b}\int_0^{\infty}\frac{\sin tx}{x}\,dx\,dt\\&=\int_a^b \frac{\pi}{2}\,dt\\&=\frac{(b-a)\pi}{2}\end{aligned} $$