I have the integral:
$$ \int_{r_1}^{r_2} \int_0^\infty f(m)J_0(mr)\,\text{d}m \,\text{d}r \tag{1}\label{1}$$
where $f(m)$ is a known but complex function, not integrable in closed form, but still continuous and non-increasing in $(0, \infty)$ interval. I need to evaluate this integral numerically, frequently. My current approach is using the Hankel transform like this:
$$ \mathcal{H}_0(r) = \int_0^\infty \frac{f(m)}{m}mJ_0(mr) \, \text{d}m, \tag{2} \label{2}$$ and then: $$ \int_{r_1}^{r_2} \mathcal{H}_0(r) \, \text{d}r. \tag{3}\label{3}$$ Both $\eqref{2}$ and $\eqref{3}$ can be evaluated numerically with proper quadrature rules, however this needs the Hankel tranform to be performed many times which becomes very inefficient. Thus my question is: can i represent integral \eqref{1} in other form, which could be more efficient, possibly not involving frequent computation of Hankel transform?
Update: $f(m)$ is the rational function like this example:
$$ f(m) = - \frac{0.2525 \left(0.2302 m^{3} e^{6 m} - 0.3568 m^{2} e^{8 m} + 0.3568 m^{2} e^{4 m} + 0.4918 m e^{10 m} + 1.0 m e^{6 m} + 0.0216 m e^{2 m} - 0.7623 e^{12 m} - 0.0243 e^{8 m} + 0.1021 e^{4 m} + 0.0335\right)}{0.0581 m^{4} e^{6 m} + 0.1242 m^{2} e^{10 m} + 0.0901 m^{2} e^{8 m} + 1.0 m^{2} e^{6 m} + 0.0901 m^{2} e^{4 m} + 0.0055 m^{2} e^{2 m} + 0.1925 e^{12 m} + 0.1584 e^{10 m} + 0.0472 e^{8 m} + 0.1225 e^{6 m} + 0.0276 e^{4 m} + 0.007 e^{2 m} + 0.0085} $$
Regards, Marek.