Fourier-Bessel series coefficients When finding the coefficients of a Fourier-Bessel series, the Bessel functions satisfies, for $k_1$and $k_2$ both zeroes of $J_n(t)$,  the orthogonality relation given by:
$$\int_0^1 J_n(k_1r)J_n(k_2r)rdr = 0, (k_1≠k_2)$$
and for $k_1 = k_2 = k$:
$$\int_0^1 J_n^2(kr)rdr = \frac12J_n^{'2}(k)$$
I understand how to get the first result since the Bessel's equation can be interpreted as a Sturm-Liouville problem, but how can I show the second one?
 A: Given a Sturm–Liouville-type differential equation
$$ -(pu')'+qu=\lambda w u, $$
(boundary conditions are not important yet) we can look at the following combination of two solutions $u_i$ with eigenvalues $\lambda_i$, which bears a resemblance to the Wronskian:
$$ v = p(u_1'u_2-u_2'u_1). $$
Differentiating and using the differential equations,
$$ v' = (pu_1')'u_2 + pu_1'u_2'-pu_2'u_1' - (pu_2')'u_1 = -(\lambda_1 w-q)u_1u_2+(\lambda_2 w-q)u_2u_1 \\
= -(\lambda_1-\lambda_2)wu_1u_2. $$
Hence we obtain the indefinite integral
$$ \int w u_1 u_2 = -p\frac{u_1'u_2 - u_2'u_1}{\lambda_1-\lambda_2}. \tag{1} $$
Now specialise to Bessel's equation, which can be interpreted as a Sturm–Liouville equation with $p=r$ and $w=r$ with solutions $J_n(kr)$ that are regular at $0$ with eigenvalue $k^2$. To satisfy the boundary condition at $1$, we need $J_n(k)=0$.
To obtain orthogonality, take $k_1 \neq k_2$ solutions to $J_n(k)=0$, and then
$$ \int_0^1 J_n(k_1r)J_n(k_2r) r \, dr = -\frac{J_n'(k_1)J_n(k_2)-J_n'(k_2)J_n(k_1)}{k_1^2-k_2^2} = 0. $$
To obtain the condition for $k_1=k_2$, let $k_1$ satisfy $J_n(k_1)=0$, and $k_2=k_1+\varepsilon$. Now it's useful that (1) holds for functions that don't satisfy the boundary conditions, because we can write
$$ \int_0^1 J_n(k_1r)J_n((k_1+\varepsilon)r) r \, dr = -\frac{J_n'(k_1)J_n(k_1+\varepsilon)}{k_1^2-(k_1+\varepsilon)^2}. $$
Then $ J_n(k_1+\varepsilon) = 0 + \varepsilon J_n'(k_1) +O(\varepsilon^2) $ and $ k_1^2-(k_1+\varepsilon)^2 = -2k_1\varepsilon + O(\varepsilon^2) $, so taking $\varepsilon \to 0$ gives the result.
