I am trying to find the Fourier-Bessel series for the function $$f(x)=1-x\ \ \text{for} \ \ 0<x<1.$$
The Fourier-Bessel series has the form $$f(x)=\sum_{n=1}^{\infty} A_nJ_v(k_nx), \ \ 0<x<l, \ \ \text{where} \ \ A_n=\frac{2}{J_{v+1}(k_n)^2}\int_{0}^{1} (1-x)J_v(k_nx)x \ dx.$$ I have tried to solve $A_n$ as follows. \begin{align} A_n=\frac{2}{J_{v+1}(k_n)^2}\int_{0}^{1}xJ_v(k_nx) \ dx-\frac{2}{J_{v+1}(k_n)^2}\int_{0}^{1} x^2J_v(k_nx) \ dx.. \end{align} but I am bit unsure of how to proceed, as I do not know what the value of $v$ is. I know somewhere I will need to use the identity $$(k_nx)^vJ_v(k_nx)=\int (k_nx)^vJ_{v-1}(k_nx) \ dx,$$ but I am a little bit stuck.
Many thanks kind people (I am new to this site)
Fourier-Bessel transform originates from a 2D transform, where you need two indices and $v$ enumerates the angular part. If your problem is part of a 2D problem, then you get the answer there. Otherwise, consider this: Bessel functions $J_v$ behave as $r^v$ around the origin, so $v=0$ is the only one that is capable of describing nonzero value at $r=0$ smoothly. If you do this for $v=1$ or higher, it will develop your function into something that has a discontinuous jump at the origin (similar to what happens if you expand a nonzero constant value into a sine series: you get a square wave).
So, every $v$ will describe a different orthogonal basis, but with different boundary condition at the origin, so $v=0$ is a natural choice for you.