Question
What are coefficients of the Fourier-Bessel expansion of the constant function 1?
My Solution
Define $J_n:\mathbb{R}\to \mathbb{R}$ as the $n$'th order Bessel function of the first kind. Define $\alpha_i\in \mathbb{R}^+$ as the $i$'th positive root of $J_0$.
Let $$z_i = \int_{r = 0}^1 J_0(\alpha_i r) r\, \mathrm{d} r.$$ Mathematica says that
$$z_i = \frac{J_1(\alpha_i)}{\alpha_i}.$$ Is this as close to "closed form" as I can get?
I have looked in the Wolfram functions page for identities of this type but nothing seemed to fit the bill. Any ideas?
Recall the identities
$$\begin{align} xJ_0(x)&=xJ_1'(x)+J_1(x) \tag 1\\\\ J_0'(x)&=-J_1(x) \tag 2 \end{align}$$
Then, we have from $(1)$ and $(2)$
$$xJ_0(x)=\left(xJ_1(x)\right)' \tag 3$$
Using $(3)$ reveals
$$\begin{align} \int_0^1J_0(ar)\,r\,dr&=\frac{1}{a}\int_0^1\frac{d\,\left(rJ_1(ar)\right)}{dr}\,dr\\\\ &=\frac{1}{a}J_1(a) \end{align}$$
Letting $a=\alpha_i$ recovers the result sought in the OP.
As requested in comment, the evaluation of the integral
$$J_1^2(\alpha_i)=2\int_0^1J_0^2(\alpha_ir)d\,dr$$
is easily facilitated by the identity
$$\frac{d}{dx} \left(\frac12 x^2 \left(J_0^2(x)+J_1^2(x)\right)\right)=xJ_0^2(x) $$
and the fact the $J_0(\alpha_i)=0$