Normalisation of Bessel functions

Question

I've done the integration by parts and obtained

$$ \frac{-1}{\alpha^2} \int z^2 J J'$$

but I have no idea how to use Bessel's equation to simplify this as it only appears to get far more complicated.

Answer 1

Call the integral under inspection for $I$, and let all $J=J_n$. The substitution implies that we are bound to show that $$ \frac{1}{\alpha^2}\int_0^\alpha z(J(z))^2\,dz=\frac{1}{2}(J'(\alpha))^2. $$ I think it is more intuitive to start the work from the differential equation. First, we have the Bessel equation $$ z^2J''+zJ'+(z^2-n^2)J=0, $$ which can be written $$ z(zJ')'=(n^2-z^2)J. $$ Now multiply by $2J'$ to get $$ 2(zJ')'zJ'=(n^2-z^2)2JJ', $$ or $$ ((zJ')^2)'=(n^2-z^2)(J^2)' $$ Integrating from $0$ to $\alpha$, and then by parts, we get $$ \begin{aligned} \alpha^2(J'(\alpha))^2&=\int (n^2-z^2)(J^2)'\,dz\\ &=\bigl[(n^2-z^2)(J(z))^2\bigr]_0^{\alpha}+2\int_0^\alpha z(J(z))^2\,dz\\ &=2\int_0^\alpha z(J(z))^2\,dz \end{aligned} $$ and we are done. Note that in the last step, we use that $(n^2-0^2)=0$ if $n=0$ and that $J_n(0)=0$ if $n>0$.