How do I show that:
$$\int_0^1{x[J_n(\alpha x)]^2dx}=\frac1 2[J_n'(\alpha)]^2$$
where $\alpha$ is a zero of the Bessel function?
Switching to $z=\alpha x$ gives:
$$\frac1 {\alpha^2}\int_0^\alpha{z[J_n(z)]^2dz}$$
But here I'm stuck. I can use integration by parts, and replace $J_n''$ from the Bessel equation when it comes up. But that introduces an additional $z^2$ term, so I get no closer to proving the equality.
Let $u_1$ and $u_2$ satisfy the Sturm–Liouville equations $$ -(pu_i')'+qu_i = \lambda_i w u_i $$ (we assume nothing about boundary conditions). Then $$ \frac{d}{dx} p(u_1'u_2-u_2'u_1) = (pu_1')'u_2-(pu_2')'u_1 = (q-\lambda_1 w)u_1u_2-(q-\lambda_2 w)u_1u_2 = (\lambda_2-\lambda_1)w u_1 u_2, $$ so $$ \int w u_1 u_2 \, dx = -p\frac{u_1'u_2-u_2'u_1}{\lambda_1-\lambda_2}. $$
In particular, $y=J_{\nu}(\alpha x)$ satisfies the equation $$ -(xy')' + \frac{\nu^2}{x} y = \alpha^2 x y, $$ which gives us the indefinite integral $$ \int x J_{\nu}(\alpha x) J_{\nu}(\beta x) \, dx = -x\frac{\alpha J_{\nu}'(\alpha x) J_{\nu}(\beta x) - \beta J_{\nu}'(\beta x) J_{\nu}(\alpha x) }{\alpha^2-\beta^2}, $$ and so for $\nu \geq -1/2$, the definite integral $$ \int_0^1 x J_{\nu}(\alpha x) J_{\nu}(\beta x) \, dx = -\frac{\alpha J_{\nu}'(\alpha) J_{\nu}(\beta) - \beta J_{\nu}'(\beta) J_{\nu}(\alpha) }{\alpha^2-\beta^2}, $$ since we always have either $xJ_{\nu}(x)J'_{\nu}(x) \to 0$ as $x \to 0$ in this case. Suppose that $\alpha$ is a zero of $ J_{\nu}$, and this reduces to $$ \int_0^1 x J_{\nu}(\alpha x) J_{\nu}(\beta x) \, dx = -\frac{\alpha J_{\nu}'(\alpha) J_{\nu}(\beta)}{\alpha^2-\beta^2}. $$ If we now let $\beta \to \alpha$ and expand $$ J_{\nu}(\beta) = (\beta-\alpha)J_{\nu}'(\alpha) + O((\beta-\alpha)^2), $$ we obtain $$ \int_0^1 x [J_{\nu}(\alpha x)]^2 \, dx = \frac{1}{2}[J_{\nu}'(\alpha)]^2$$ as desired.