Your formula would not work just by changing the domain of integration. In the same vein there is another useful basin given by the Fourier Bessel series. The idea is to go in polar coordinates and diagonalise the Laplacian by separation of variables. You write:
$$
f(r,\phi) = \sum_{m,n\in\mathbb N^2} c_{mn}e^{im\phi}J_m(u_{mn}r)
$$
with $u_{mn}$ the $n$th root of $J_m$. You can calculate the coefficients with:
$$
c_{mn} = \frac{1}{\pi J_{m+1}(u_{mn})}\int_0^{+\infty}rdr\int_0^{2\pi}d\phi e^{-im\phi}J_m(u_{mn}r)f(r,\phi)
$$
You can check that the basis functions $e_{mn}=e^{im\phi}J_m(u_{mn}r)$ are orthogonal and are eigenvectors of the Laplacian:
$$\Delta=\partial_r^2+\frac{1}{r}\partial_r+\frac{1}{r^2}\partial_\phi^2$$
Your formula would not work just by changing the domain of integration. In the same vein there is another useful basin given by the Fourier Bessel series. The idea is to go in polar coordinates and diagonalise the Laplacian by separation of variables. You write: $$ f(r,\phi) = \sum_{m,n\in\mathbb N^2} c_{mn}e^{im\phi}J_m(u_{mn}r) $$ with $u_{mn}$ the $n$th root of $J_m$. You can calculate the coefficients with: $$ c_{mn} = \frac{1}{\pi J_{m+1}(u_{mn})}\int_0^{+\infty}rdr\int_0^{2\pi}d\phi e^{-im\phi}J_m(u_{mn}r)f(r,\phi) $$
You can check that the basis functions $e_{mn}=e^{im\phi}J_m(u_{mn}r)$ are orthogonal and are eigenvectors of the Laplacian: $$\Delta=\partial_r^2+\frac{1}{r}\partial_r+\frac{1}{r^2}\partial_\phi^2$$
Hope this helps.