Integral ${\large\int}_0^\infty\big(2J_0(2x)^2+2Y_0(2x)^2-J_0(x)^2-Y_0(x)^2\big)\,dx$
I stated an interesting thing in the comments (wisely explored by Random Variable), but it is definitely not the fastest way to go.
For the glory of Italian mathematicians, Frullani's theorem is the way (again):
$$ \int_{0}^{+\infty}\frac{x\, J_0(x)^2-2x\, J_0(2x)^2}{x}\,dx = (-\log 2)\cdot\underset{x\to+\infty}{\widetilde{\lim}} x\cdot J_0(x)^2 = \color{red}{-\frac{\log 2}{\pi}} $$ and the argument is just the same for the $Y_0$ function.
The wide tilde-limit has to be intended as a Cesàro mean: $$\underset{x\to+\infty}{\widetilde{\lim}}f(x) = \lim_{x\to +\infty}\frac{1}{x}\int_{x}^{2x}f(z)\,dz.$$
Moreover, provided that the integral converges, the previous result holds unchanged also by replacing $J_0$ with $J_\nu$ for any $\nu\geq 0$, since: $$ J_\nu(z) = \sqrt{\frac{2}{\pi z}}\left(\cos \left(z-\frac{\nu\pi}{2}-\frac{\pi}{4}\right)+O\left(\frac{1}{z}\right)\right) $$ for any large $z\in\mathbb{R}^+$.
Somewhat similar to this answer, we can use Ramanujan's master theorem to show that $$\int_{0}^{\infty}\big(2J_0(2x)^2-J_0(x)^2\big)\,dx = \frac{\ln 2}{\pi}. $$
As $x \to \infty$, $$\big(2J_0(2x)^2-J_0(x)^2\big) \sim \frac{\sin (4x)-\sin (2x)}{\pi x}. \tag{1}$$ So the integral does indeed converge (but not absolutely).
The hypergeometric representation of the square of the Bessel function of the first kind of order zero is $$J_{0}(z)^{2} = \, _1F_2\left(\frac{1}{2}; 1, 1; -z^{2} \right). $$
So for $a>0$ and $0 < s < 1$, the Mellin transform of $J_{0}(ax)^{2}$ is $$ \begin{align} \int_{0}^{\infty} x^{s-1} J_{0}(ax)^{2} \,dx &= \int_{0}^{\infty} x^{s-1} \, _1F_2\left(\frac{1}{2}; 1, 1; -(ax)^{2} \right) \, dx \\ &= \frac{1}{2a^{s}}\int_{0}^{\infty} u^{s/2-1} \, _1F_2\left(\frac{1}{2}; 1, 1; -u \right) \, du \\ &= \frac{1}{2a^{s}} \Gamma\left(\frac{s}{2} \right) \frac{\Gamma \left(\frac{1}{2}-\frac{s}{2} \right) \Gamma(1)^{2} }{\Gamma\left(\frac{1}{2}\right) \Gamma\left(1- \frac{s}{2} \right)^{2}} \\ &= \frac{1}{2a^{s}} \Gamma\left(\frac{s}{2} \right) \frac{\Gamma \left(\frac{1}{2}-\frac{s}{2} \right)}{\sqrt{\pi} \, \Gamma\left(1- \frac{s}{2} \right)^{2}} . \end{align}$$
Therefore, assuming we can bring the limit inside the integral, $$ \begin{align} \int_0^\infty\big(2J_0(2x)^2-J_0(x)^2\big)\,dx &= \frac{1}{\pi} \lim_{s \to 1^{-}} \left(\frac{1}{2^{s}}- \frac{1}{2} \right) \Gamma \left(\frac{1}{2} - \frac{s}{2} \right) \\ &= \frac{1}{\pi} \lim_{s \to 0^{-}} \left( \frac{1}{2^{s+1}} - \frac{1}{2} \right)\Gamma \left(- \frac{s}{2} \right) \\ &= \frac{1}{\pi} \lim_{s \to 0^{-}} \left(- \frac{\ln 2}{2} s + \mathcal{O}(s^{2}) \right) \left(- \frac{2}{s} + \mathcal{O}(1) \right) \\ &= \frac{\ln 2}{\pi}. \end{align}$$
The same approach seemingly also shows that $$\int_{0}^{\infty}\big(2J_v(2x)^2-J_v(x)^2\big)\,dx = \frac{\ln 2}{\pi} \, , \quad v> - \frac{1}{2}. $$
$(1)$ https://en.wikipedia.org/wiki/Bessel_function#Asymptotic_forms