I read in a book that
$$\int_0^1x[J_n(\alpha x)]^2dx = \frac{1}{2}[J_n'(\alpha)]^2$$
where $\alpha$ is a zero of $J_n$ and where $J_n$ is a solution to Bessels equation. The book gives a hint telling me to use the substitution $z=\alpha x$ and then integrating by parts and then use the fact that $J_n$ solves Bessels equation.
I tried to use the hint to prove this, but end up going in circles when Integrating by parts.
Changing as indicated $z=\alpha x$ and integrating by parts, \begin{align} I&=\int_0^1x[J_n(\alpha x)]^2\,dx \\ &= \frac{1}{\alpha^2}\int_0^\alpha z[J_n(z)]^2\,dz\\ &=-\frac{1}{\alpha^2}\int_0^\alpha z^2J_n(z)J'_n(z)\,dz \end{align} Now, from the differential equation for the Bessel functions, \begin{equation} z^2J_n(z)=n^2J_n(z)-zJ'_n(z)-z^2J''_n(z) \end{equation} Plugging this expression into the last integral, it comes \begin{align} I&=-\frac{1}{\alpha^2}\int_0^\alpha J'_n(z)\left[n^2J_n(z)-zJ'_n(z)-z^2J''_n(z)\right]\,dz\\ &=-\frac{1}{\alpha^2}\left[\frac{n^2}{2}J_n^2(z)-\int_0^\alpha \left( zJ'_n(z)+z^2J''_n(z) \right)J_n(z)\,dz\right]\\ &=\frac{n^2}{2\alpha^2}J_n^2(0)+\frac{1}{\alpha^2}\int_0^\alpha \left( zJ'_n(z)+z^2J''_n(z) \right)\,dz\\ &=\frac{n^2}{2\alpha^2}J_n^2(0)+\frac{1}{2\alpha^2}\int_0^\alpha \frac{d}{dz}\left( z^2 [J'_n(z)]^2\right)\,dz\\ &=\frac{1}{2}[J_n'(\alpha)]^2 \end{align} In the last expression we used that $J_n(0)=0$ if $n\neq 0$.