TITLE: Double integral with Hankel transform
QUESTION [10 upvotes]: Let's say we have a double integral in the following form:
$$I=\int_0^\infty \int_0^\infty f(x) g(y) J_0(xy) x y dx dy $$
Using the definition of the Hankel transform, we can write:
$$I=\int_0^\infty  F(y) g(y)  y dy=\int_0^\infty  G(x) f(x)  x dx$$
Or, using the self-inverse property of the transform:
$$I=\int_0^\infty \int_0^\infty F(x) G(y) J_0(xy) x y dx dy $$
I wasn't able to find my functions of interest in the tables of direct or inverse Hankel transforms. Which is why I decided to ask a general question:

Is there a way to simplify such an integral if we don't know the exact Hankel transform for either function?

I know this is a long shot, but maybe there are some identities I don't know which could help.

Just in case it's important, my functions are:
$$f(x)=\frac{e^{-ax}}{x^2+b^2} \\ g(y)=e^{-y} L_l \left( \frac{2L+1}{l+L+1} y \right) L_L \left( \frac{2l+1}{l+L+1} y \right)$$
Where $L_l$ are Laguerre polynomials.
So this could also be related to Laplace transform. Though I have more hopes for Hankel, as one of the integrals is already a result of Laplace transform and going back would be counter-intuitive.
But I'm interested in the general case too.

Probably wouldn't help, but there's an interesting connection between Laguerre polynomials and Bessel functions, in particular:
$$L_n(z)= \frac{2}{n!} e^z z^{n+1} \int_0^\infty e^{-z t^2} t^{2n+1} J_0(2zt) dt$$
More useful integrals I found in G-R: 
$$\int_0^\infty \frac{x \sin (ax)}{x^2+b^2} J_0(y x) dx=\frac{\pi}{2} e^{-ab} I_0 (b y), \qquad y \leq a$$
$$\int_0^\infty \frac{x \cos (ax)}{x^2+b^2} J_0(y x) dx= \cosh (a b) K_0(b y), \qquad y \geq a $$
Heck if I know how to convert these results to the exponential form, as they work for different ranges of $y$.

REPLY [6 votes]: $\textbf {Edition of 05.12.2018}$
$$\color{brown}{\textbf{1. Integral representation of the Bessel function}}$$
Let us consider another way using the integral representation of the Bessel function $(69)$

$$J_0(z)=\dfrac1\pi\int\limits_0^\pi \cos(z\sin\theta)\,\mathrm d\theta=\dfrac2\pi\int\limits_0^{\pi/2} \cos(z\sin\theta)\,\mathrm d\theta.\tag{1.1}$$

Substitution $\sin\theta = t$ transforms it to
$$J_0(z) = \dfrac2\pi\int\limits_0^1\dfrac{\cos(zt)}{\sqrt{1-t^2}}\mathrm dt.\tag{1.2}$$
This leads to the formula
$$I = \dfrac2\pi\int\limits_0^1\dfrac{C(t)}{\sqrt{1-t^2}}dt,\tag{1.3}$$
where 
$$C(t) = \int\limits_0^\infty \int\limits_0^\infty f(x)g(y)xy\cos(xyt)\,\mathrm{dx}\,\mathrm{dy}.\tag{1.4}$$
Easy to see that formulas $(1.3)-(1.4)$ allow to replace Bessel function to cosine one.
$$\color{brown}{\textbf{2. Testing on $1D$ Hankel transform}}$$

By the definition, $$F(q)=\mathcal H(f(x))= 2\pi\int\limits_0^\infty
xf(x)J_0(2\pi qx)\,\mathrm dx.\tag{2.1}$$

Identity $(1.2)$ allows to write $(2.1)$ in the form of
$$F(q)= 4\int\limits_0^1\int\limits_0^\infty \dfrac{xf(x)\cos(2\pi qxt)}{\sqrt{1-t^2}}\,\mathrm dx\,\mathrm dt
 = 4\int\limits_0^1 \dfrac{\Phi(2\pi qt)}{\sqrt{1-t^2}}\mathrm dt,\tag{2.2}$$
where
$$\Phi(p) = \int\limits_0^\infty xf(x)\cos(px)\,\mathrm dx.\tag{2.3}$$
Let 
$$f(x) = \dfrac {e^{-sx}}x,$$
then
$$F(q)=\mathcal H\left(\dfrac {e^{-sx}}x\right) = 2\pi \int\limits_0^\infty e^{-sx}J_0(2\pi qx)\mathrm dx = 2\pi\mathcal L(J_0(2\pi qx))
=\dfrac{2\pi}{\sqrt{s^2+4\pi^2q^2}}.\tag{2.4}$$
On the other hand,
$$\Phi(p) = \int\limits_0^\infty e^{-sx}\cos(px)\,\mathrm dx = \mathcal L(\cos(px)) = \dfrac s{s^2+p^2}\tag{2.5}$$
and
$$F(q) = 4\int\limits_0^1 \dfrac{\Phi(2\pi qt)}{\sqrt{1-t^2}}\mathrm dt
 = 4\int\limits_0^1 \dfrac{s}{s^2+4\pi^2q^2t^2}\dfrac{\mathrm dt}{\sqrt{1-t^2}} = \dfrac{2\pi}{\sqrt{s^2+4\pi^2q^2}}\tag{2.6}$$
(see also Wolfram Alpha).
Results $(2.4)$ and $(2.6)$ are the same. 
Testing is successful.
$$\color{brown}{\textbf{3. Integration via the given function $f(x)$}}$$
For the given function $f(x),$
$$\int\limits_0^\infty\dfrac{xe^{-kx}}{x^2+b^2}\,\mathrm dx = \int\limits_0^\infty\dfrac{\dfrac xbe^{-kb\frac xb}}{\left(\dfrac xb\right)^2+1}\,\mathrm d\dfrac xb = \int\limits_0^\infty\dfrac{xe^{-kbx}}{x^2+1}\,\mathrm dx = F_1(kb),\tag{3.1}$$
where
$$F_1(s)=\left(\dfrac\pi2 - \mathrm{Si}(s)\right)\sin(s) - \dfrac12\mathrm{Ci}(s)\cos(s))\tag{3.2}$$
(see also Wolfram Alpha).
Then
$$\int\limits_0^\infty\dfrac{xe^{-ax}\cos(ytx)}{x^2+b^2}\,\mathrm dx = \dfrac{F_1(-ab+iybt)+F_1(-ab-iybt)}2,$$
$$C(t) = \int\limits_0^\infty yg(y)\dfrac{F_1(-ab+iybt)+F_2(-ab-iybt)}2\,\mathrm{dy}.\tag{3.3}$$
And the next step looks too complex.
$$\color{brown}{\textbf{4. Integration via the given function $g(y)$}}$$
For the given function $g(y),$
$$\int\limits_0^\infty yg(y)\cos(xty)\,\mathrm dy = \int\limits_0^\infty ye^{-y} L_l \left( \frac{2L+1}{l+L+1} y \right) L_L \left( \frac{2l+1}{l+L+1} y \right)\cos(xty)\,\mathrm dy = \dfrac{G(1-ixt)+G(1+ixt)}2,\tag{4.1}$$
where 
$$G(s) = \int\limits_0^\infty ye^{-sy} L_l(y+py) L_L(y-py)\,\mathrm dy,\tag{4.2}$$
or
$$G(s) = -\dfrac{\mathrm d}{\mathrm ds}\mathcal L\bigl(L_l(y+py) L_L(y-py)\bigr),\tag{4.3}$$
where
$$p = \frac{L-l}{l+L+1}\tag{4.4},$$

Is known formula for the convolution 
   $$\int\limits_0^y L_m(t)L_n(y-t)\,\mathrm
 dt = \int\limits_0^y L_{m+n}(t)\,\mathrm dt =
 L_{m+n}(y)-L_{m+n+1}(y)\tag{4.5}$$
  But I can't apply it for obtained kind of the integrals.

However, for the given values $l$ and $L$ the integral $(4.2)$ can be calculated and presented in the form of
$$G(s) = \dfrac {P(s)}{s^{L+l+1}},\tag{4.6}$$
wherein degree of the polynomial $P(s)$ is less than $L+l+1.$
Then
$$C(t) = \dfrac12 \int\limits_0^\infty xf(x)\left(\dfrac{P(1-ixt)}{(1-ixt)^{L+l+1}} + \dfrac{P(1+ixt)}{(1+ixt)^{L+l+1}}\right)\,\mathrm dx,\tag{4.7}$$
wherein
$$xf(x)\left(\dfrac{P(1-ixt)}{(1-ixt)^{L+l+1}} + \dfrac{P(1+ixt)}{(1+ixt)^{L+l+1}}\right)=\left(\dfrac{C_1(t)x+C_2(t)}{x^2+b^2}+\sum_{k=1}^{L+l+1}\dfrac{A_k(t)}{(1-ixt)^k}+\sum_{k=1}^{L+l+1}\dfrac{B_k(t)}{(1+ixt)^k}\right)e^{-ax}.$$
Let us calculate the integrals
$$\int\limits_0^\infty\dfrac{e^{-kx}}{x^2+b^2}\,\mathrm dx = \dfrac1b \int\limits_0^\infty\dfrac{e^{-kb\frac xb}}{\left(\dfrac xb\right)^2+1}\,\mathrm d\dfrac xb = \dfrac1b \int\limits_0^\infty\dfrac{e^{-kbx}}{x^2+1}\,\mathrm dx = \dfrac1b F_0(kb),\tag{4.8}$$
where
$$F_0(s)=\mathrm{Ci}(s)\sin(s) + \left(\dfrac\pi2-\mathrm{Si(s)}\right)\cos(s)\tag{4.9}$$
(see also Wolfram Alpha),
$$\int\limits_0^\infty\dfrac{e^{-kx}}{1\pm ixt} = \dfrac 1t \int\limits_0^\infty\dfrac{e^{-\frac kt xt}}{1\pm ixt}\,\mathrm d(xt) = \dfrac 1t \int\limits_0^\infty\dfrac{e^{-\frac kt x}}{1\pm ix}\,\mathrm dx = \dfrac 1t \int\limits_0^\infty\dfrac{1\mp ix}{x^2+1}e^{-\frac kt x}\,\mathrm dx,$$
$$\int\limits_0^\infty\dfrac{e^{-kx}}{1\pm ixt} = \dfrac1t F_0\left(\dfrac kt\right) \mp \dfrac it F_1\left(\dfrac kt\right),\tag{4.10}$$
$$H_m(s)=\int\limits_0^\infty\dfrac{e^{-sx}}{(1\pm ix)^m}\,\mathrm dx
= -\dfrac1{m-1} \int\limits_0^\infty e^{-sx}\,\mathrm d(1\pm ix)^{-m+1}$$
$$= -\dfrac1{m-1} e^{-sx}(1\pm ix)^{-m+1}\bigg|_0^\infty + \dfrac s{m-1}\int\limits_0^\infty\dfrac{e^{-sx}}{(1\pm ix)^{m-1}}\,\mathrm dx
= \dfrac 1{m-1}\left(1 + sH_{m-1}(s)\right),$$
$$H_m(s)=\sum\limits_{n=0}^{m-2}\dfrac{(m-n-2)!}{(m-1)!}s^n+\dfrac1{(m-1)!}(F_0(s)\mp F_1(s)),\tag{4.11}$$
$$\int\limits_0^\infty\dfrac{e^{-kx}}{(1\pm ixt)^m}dx = \dfrac1t H_m\left(\dfrac kt\right).\tag{4.12}$$
Obtained formulas allow to calculate $C(t)$ 
in the closed form, after what remains using of $(1.3).$ 
Therefore, the main problem is to calculate $1$D integral which contains Ci and Si special functions.
These are the reasons why this approach can be more successful than Hankel transform.