I am trying to find the Fourier-Bessel series for $f(x) = x$, $0 < x < 2$, with respect to the orthogonal set $\{ J_1 (k_n(x)) \}$, where $k_n$ is the $n^{th}$ positive root of the equation $J_1(2k) = 0$.
I am told that
$$\int_0^c x [J_\alpha (k_n(x)]^2 \ dx = \dfrac{c^2}{2} [J_{\alpha + 1}(k_n c)]^2,$$
where $k_n$ is a root of $J_{\alpha}(kc) = 0$.
I am also told that
$$\dfrac{d}{dx} [x^{\alpha} J_{\alpha}(x)] = x^{\alpha} J_{\alpha - 1}(x)$$
What I Know About Fourier-Bessel Series:
For any fixed $m$, the functions
$$\left\{ J_m \left( \dfrac{\mu_n^{(m)}x}{a} \right) \right\}_{n = 1}^\infty$$
are orthogonal w.r.t. the weight function $w(x) = x$ on $0 < x < a$. Here, $\mu_n^{(m)}$ is the $n^{th}$ zero of $J_m(\mu)$.
The Fourier-Bessel series for a function $f(x)$ defined on $(0, a)$ is
$$f(x) = \sum_{n = 1}^\infty c_n J_m \left( \dfrac{\mu_n^{(m)}x}{a} \right)$$
I have absolutely no experience with Fourier-Bessel series or Bessel functions in general.
I would greatly appreciate it if people could please take the time to explain this problem.
Note that there is no $k_n(x)$, $k_n$ depends only on $n$. For convenience, let's normalize the basis. Let $$e_n(x) = \frac 1 {\sqrt 2 \,J_2(2 k_n)} J_1(k_n x), \\ J_1(2 k_n) = 0, \;n \in \mathbb N.$$ Then, with the inner product defined as $$(f, g) = \int_0^2 x f(x) g(x) dx,$$ ${e_n}$ form an orthonormal basis, $(e_m, e_n) = \delta_{mn}$. The expansion of $f$ in this basis has the form $$\sum_n (f, e_n) e_n.$$ To compute the coefficients $(f, e_n)$ for $f(x) = x$, we use the fact that $$\int x^2 J_1(a x) dx = \frac {x^2 J_2(a x)} a,$$ from which $$(f, e_n) = \frac 1 {\sqrt 2 \,J_2(2 k_n)} \int_0^2 x^2 J_1(k_n x) dx = \frac {2 \sqrt 2} {k_n}.$$