How to solve POTW #290?

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Here is this week's POTW:

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Show that $$\int_0^\infty \left[\frac{J_1(t)}{t}\right]^2\, dt = \frac{4}{3\pi}$$

where $J_1$ is the Bessel function of the first kind of order one.
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No one answered this week's problem. You can read my solution below.

Spoiler

Note $$J_1(t) = \frac{1}{2\pi}\int_{-\pi}^{\pi} e^{-i\theta + it\sin(\theta)}\, d\theta$$ so that $$[J_1(t)]^2 = \frac{1}{4\pi^2} \int_{-\pi}^\pi \int_{-\pi}^\pi e^{-i(\theta + \phi) + it(\sin(\theta) + \sin(\phi))}\, d\theta\, d\phi = \frac{1}{2\pi^2} \int_0^\pi \int_{-\pi}^\pi e^{-2iu + 2it\sin u \cos v}\, du\, dv = \frac{1}{\pi}\int_0^\pi J_2(2t\cos v)\, dv = \frac{2}{\pi}\int_0^{\pi/2} J_2(2t\cos v)\, dv$$ Thus
$$\int_0^\infty \left[\frac{J_1(t)}{t}\right]^2\, dt = \frac{2}{\pi}\int_0^{\pi/2} \int_0^\infty\frac{J_2(2t\cos v)}{t^2}\, dt\, dv = \frac{2}{\pi}\int_0^{\pi/2} 2\cos v\, dv \int_0^\infty \frac{J_2(t)}{t^2}\, dt = \frac{2}{\pi}\cdot 2\cdot \frac{1}{3} = \frac{4}{3\pi}$$