I am looking for a reference for tables of Hankel/spherical Bessel tranforms. In particular, I'm trying to calculate transforms like
\begin{align}
f_{LM}(r) & = i^L \sqrt{\frac{2}{\pi}} \int_0^\infty k^2\ dk\ j_L(k r)\ \tilde f_{LM}(k)
\\
& = i^L \frac{1}{\sqrt{r}} \int_0^\infty k\ dk\ \sqrt{k} J_{L+1/2}(kr)\ \tilde f_{LM}(k),
\end{align}
where $j_L(kr)$ is a spherical Bessel function and $J_\nu(kr)$ is a (cylindrical) Bessel function. These are the transforms associated with Fourier transforming a function which has been decomposed into multipoles:
\begin{gather}
\begin{aligned}
f(r,\theta_r,\phi_r) &= \sum_{LM} f_{LM}(r) Y_{LM}(\theta_r,\phi_r), & \tilde{f}(k,\theta_k,\phi_k)&=\sum_{LM}\tilde{f}(k) Y_{LM}(\theta_k,\phi_k),
\end{aligned} \\
f(r,\theta_r,\phi_r) = \int \frac{d^3k}{(2\pi)^{3/2}} e^{i \mathbf{k}\cdot\mathbf{r}} \tilde{f}(k,\theta_k,\phi_k).
\end{gather}
I have tried using Mathematica's HankelTransform
function, but it seems to often miss the singular pieces. For instance, with $L=0$, both
$$\tilde f_{00}(k) = \frac{1}{k^2+m^2} \quad \text{and} \quad \tilde f_{00}(k) = \frac{-k^2/m^2}{k^2+m^2}$$
give a transform of
$$f_{00}(r) = \sqrt{\frac{\pi}{2}}\frac{e^{-m r}}{r},$$
while the second function should give
$$f_{00}(r) = \sqrt{\frac{\pi}{2}}\frac{e^{-m r}}{r} - \sqrt{\frac{\pi}{2}}\frac{\delta(r)}{m^2 r^2}.$$
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1$\begingroup$ Tables of Bessel Transforms by Fritz Oberhettinger. $\endgroup$– A rural readerCommented Apr 14, 2023 at 20:30
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$\begingroup$ @Aruralreader I just took a look at that -- it seems only to include transforms where the integral actually converges. I guess I'm looking for a reference that goes further and includes pairs with (tempered?) distributions. I added distribution-theory as a tag. $\endgroup$– kc9judCommented Apr 14, 2023 at 21:19
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$\begingroup$ There should be no $1/r^2$ in your last equation, right? $\endgroup$– LL 3.14Commented Apr 15, 2023 at 8:29
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$\begingroup$ @LL3.14 I believe there should be -- without it, the inverse transform (of that term) is $$\tilde f_{00}(k) = -\frac{1}{m^2} \int_0^\infty r^2 dr\ j_0(kr) \delta(r) = 0,$$ so it has no effect. If you include the $1/r^2$, that term adds a contribution $$\tilde f_{00}(k) = -\frac{1}{m^2} \int_0^\infty dr\ j_0(kr) \delta(r) = -\frac{1}{m^2},$$ which combines with the transform of the first term to give $$\frac{1}{k^2+m^2} - \frac{1}{m^2} = \frac{-k^2/m^2}{k^2+m^2}.$$ $\endgroup$– kc9judCommented Apr 16, 2023 at 2:30
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$\begingroup$ Oh ok, I thought it was the Fourier transform at this point. In terms of Fourier transform, what do you want to compute? $\endgroup$– LL 3.14Commented Apr 16, 2023 at 12:48
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