Rayleigh's equation defines the relative intensity in an airy disk diffraction pattern. It is given as $$[1]\quad L\left(\beta q_{0}\right)=1-J_{0}^{2}\left(\beta q_{0}\right)-J_{1}^{2}\left(\beta q_{0}\right)$$ I am attempting to derive this expression, but I am stuck on a step involving evaluating a definite integral of a squared Bessel function over x.
When attempting to derive the relation from the expression for the intensity in a diffraction pattern according to Airy-Kirchhoff diffraction theory, I obtain the following expression $$[2]\quad L\left(\beta q_{0}\right)=\int_{0}^{q_{0}} J_{1}^{2}(\beta q)(d q / q) / \int_{0}^{\infty} J_{1}^{2}(\beta q)(d q / q)$$
This expression is supposed to result in Rayleigh's equation by applying the following relation:
$$[3]\quad\int_{0}^{x_{0}} J_{1}^{2}(x) \frac{d x}{x}=\frac{1}{2}\left[1-J_{0}^{2}\left(x_{0}\right)-J_{1}^{2}\left(x_{0}\right)\right]$$
This is the part I am stuck on. My knowledge of Bessel functions is very limited: I'm vaguely familiar with the integral and infinite series representations of Bessel functions and the differential equation they are a solution to. I have no idea how to approach this integral, and I'd appreciate any clarifications anybody could provide.
We can use the recurrence relations for the Bessel functions: \begin{align} J_{\nu}'\left(z\right)&=J_{\nu-1}\left(z\right)-(\nu/z) J_{\nu}\left(z\right)\\ &=-J_{\nu+1}\left(z\right)+(\nu/z) J_{\nu}\left(z\right) \end{align} respectively with $\nu=1$ and $\nu=0$ to obtain the identities \begin{align} \frac{J_1(x)}{x}&=J_0(x)-J'_1(x)\\ J'_0(x)&=-J_1(x) \end{align} and thus \begin{align} \frac{J_1^2(x)}{x}&=J_0(x)J_1(x)-J_1(x)J'_1(x)\\ &=-J_0(x)\frac{dJ_0(x)}{dx}-J_1(x)\frac{dJ_1(x)}{dx}\\ &=-\frac{1}{2}\frac{d}{dx}\left[J_0^2(x)+J_1^2(x)\right] \end{align} Finally, as $J_1(0)=0$ and $J_0(0)=1$, \begin{equation} \int_{0}^{x_{0}} J_{1}^{2}(x) \frac{d x}{x}=\frac{1}{2}\left[1-J_{0}^{2}\left(x_{0}\right)-J_{1}^{2}\left(x_{0}\right)\right] \end{equation}