Can You Crack This Tricky Integral Challenge?

[Euge]

Here is this week's POTW:

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Evaluate the integral

$$\int_{-\infty}^\infty \frac{\cos x - \cos 2x}{x^2}\, dx.$$

Justify the major steps of your calculation.
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[Euge]

No one answered this week's problem. Here is my solution.

Spoiler

The integral evaluates to $\pi$. First write

$$\int_{-\infty}^\infty \frac{\cos x - \cos 2x}{x^2}\, dx = 2\int_0^\infty \frac{\cos x - \cos 2x}{x^2}\, dx = 2\int_{0}^\infty \int_1^2 \frac{\sin tx}{x}\, dt\, dx.$$

The integral $\int_0^\infty \frac{\sin tx}{x}\, dx$ converges uniformly on $[1,2]$ since it converges (to $\frac{\pi}{2}$) on $[1,2]$ and the integrand (which we assume to have value $t$ at $x = 0$) is continuous on every rectangle $[1,2] \times [0,M]$, $M > 0$. Therefore

$$2\int_0^\infty \int_1^2 \frac{\sin tx}{x}\, dt\, dx = 2\int_1^2 \int_0^\infty \frac{\sin tx}{x}\, dx\, dt = 2\int_1^2 \frac{\pi}{2}\, dt = \pi.$$