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'm interested in the following definite integral:
$$\int_0^\infty\big(2J_0(2x)^2-J_0(x)^2+2Y_0(2x)^2-Y_0(x)^2\big)\,dx,\tag1$$
where $J_\nu$ and $Y_\nu$ are the Bessel functions of the first and the second kind.
Mathematica evaluates this integral symbolically to $\frac{\ln2}\pi$, but the result of a numerical integration looks more like $\frac{\ln4}\pi$, so I suspect the symbolic result is incorrect. 
Moreover, it looks like both components converge and make equal contribution:
$$\int_0^\infty\big(2J_0(2x)^2-J_0(x)^2\big)\,dx\stackrel?=\int_0^\infty\big(2Y_0(2x)^2-Y_0(x)^2\big)\,dx\stackrel?=\frac{\ln2}\pi,\tag2$$
but it is more difficult to check numerically, because both integrands here are oscillating (unlike $(1)$ where the integrand looks monotonic).
How can we find values of these integrals and prove them correct? Can we generalize results for values of the index $\nu$ other than $0$?

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
