There exists an orthogonality relationship with Bessel's functions of the same order $n$:
$\int_0^1 r J_{n} (\lambda_{np}r) J_n (\lambda_{nq}r) dr = 0$ if $\lambda_{np} \neq \lambda_{nq}$
and
$\int_0^1 r J_{n} (\lambda_{np}r) J_n (\lambda_{nq}r) dr = \frac{1}{2} J^2_{n+1}$ if $\lambda_{np} = \lambda_{nq}$
where $\lambda_{np}$ represents the $p^\textrm{th}$ root of the $n^\textrm{th}$ Bessel function $J_{n}$ i.e., $J_n (\lambda_{np}) = 0$. Similarly, $\lambda_{nq}$ represents the $q^\textrm{th}$ root.
However, is there an orthogonality relationship involving the roots of the derivatives of Bessel functions? For example:
$\int_0^1 r J_{n} (\lambda_{np}r) J_n (\tau_{nq}r) dr =$ ???
where $\tau_{nq}$ is the $q^\textrm{th}$ root of the first derivative of $J_n(r)$ i.e., $\frac{d J_n}{dr} = 0 $ at $r = \tau_{nq}$.
This was given as Problem 3.8 in Jackson's Electrodynamics book, 1st edition. The derivation is fairly long, but is basically the same as the derivation of your reference orthogonality integral. He gives the solution as:
$\int_0^a r J_\nu \left(\lambda _{\text{$\nu $p}} r \right) J_\nu \left(\lambda _{\text{$\nu $q}} r \right) \, dr =\frac{1}{2} a^2 \left(1-\frac{\nu ^2}{\lambda _{\text{$\nu $q}}^2}\right) J_{\nu }\left(\lambda _{\text{$\nu $q}}\right){}^2$
when $\lambda _{\text{$\nu $p}}=\lambda _{\text{$\nu $q}}$ and zero otherwise and where $\lambda _{\text{$\nu $q}}$ is the qth root of $\frac{dJ_{\nu }(r)}{dr}=0$