Ytukyg

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 $y$.

This won't answer your question, but may be of some help. Most of the information below comes from Chapter 12 of Ronald Bracewell's The Fourier Transform and Its Applications, Second Edition, Revised, 1986.

The 2-D Fourier Transform

$$ F(u,v) = mathscr{F}left{f(x,y)right}=int_{-infty}^infty int_{-infty}^infty f(x,y)space e^{-2pi i (xv+vy)} space dx dy$$

when there is circular symmetry, such that
$$f(x,y) = f(r)$$ $$r^2 = x^2+y^2$$ and thus
$$F(u,v)=F(q)$$ $$q^2 = u^2+v^2$$

can be expressed as a zeroth order Hankel Transform, with a strictly reciprocal transform

$$F(q) = mathscr{H}_0left{f(r)right} = 2pi int_0^infty f(r) space J_0(2pi qr)space rspace dr$$

$$f(r) = mathscr{H}_0left{F(q)right} = 2pi int_0^infty F(q) space J_0(2pi qr)space qspace dq$$

Bracewell provides the following theorems using the above particular Hankel Transform convention

$$begin{align*}mathscr{H}_0left{f(ar)right} &= a^{-2}Fleft(dfrac{q}{a}right) quadmbox{Similarity}\
\
mathscr{H}_0left{f(r)+g(r)right} &= Fleft(qright)+Gleft(qright) quadmbox{Addition}\
\
mathscr{H}_0left{ int_0^{2pi} int_0^infty f(r')g(R)r' space dr'space dtheta right} &= Fleft(qright)Gleft(qright) quadmbox{2D Convolution of Circularly Symmetric Functions}\
R^2 = r^2 + r'^2 -2rr'cos{theta} &\
\
int_0^infty |f(r)|^2 r space dr &= int_0^infty |F(q)|^2 q space dq quad mbox{Rayleigh}\
\
int_0^infty f(r)g^*(r) r space dr &= int_0^infty F(q)G^*(q) q space dq quad mbox{Power}\
\
mathscr{H}_0left{dfrac{d}{dr}left[rf(r)right]right} &= -dfrac{d}{dq}left[qF(q)right] quad mbox{Derivative}\
\
mathscr{H}_0left{dfrac{d}{dr}f(r)right} &= -dfrac{d}{dq}left[qmathscr{H}_0left{r^{-1}f(r)right}right] quad mbox{Derivative}\
\
mathscr{H}_0left{rdfrac{d}{dr}f(r)right} &= -q^{-1}dfrac{d}{dq}left[q^2F(q)right] quad mbox{Derivative}\
\
2pi int_0^{infty} f(r)r space dr &= F(0) quad mbox{Definite Integral}\
\end{align*}$$

There is no Shift Theorem, although you can search online for papers by Natalie Baddour that introduce a "generalized shift" and theorem for 2D Fourier Transforms with circular symmetry.

Bracewell provides the following Hankel Transform pairs that may be useful for deriving Hankel Transforms for your functions $f(x)$ and $g(y)$ above:

$$begin{align*}mathscr{H}_0left{dfrac{1}{r^2+a^2}right} &= 2pi K_0(2pi aq) \
\
mathscr{H}_0left{e^{-ar}right} &= dfrac{2pi a}{left(4pi^2q^2 + a^2right)^frac{3}{2}} \
\
mathscr{H}_0left{dfrac{e^{-ar}}{r}right} &= dfrac{2pi}{left(4pi^2q^2 + a^2right)^frac{1}{2}} \
\
mathscr{H}_0left{e^{-pi r^2}right} &= e^{-pi q^2} \
\
mathscr{H}_0left{r^2 e^{-pi r^2}right} &= left(dfrac{1}{pi} - q^2right)e^{-pi q^2} \
\
end{align*}$$

Bracewell's book also touches on 3D Fourier Transforms in cylindrical coordinates, which may or may not be useful to you.

The following online book has tables of Hankel Transforms:

https://authors.library.caltech.edu/43489/7/Volume%202.pdf

But you may need to transform the convention used. The English Wikipedia page on the Hankel Transform has a description on how to transform the convention.

Good luck.

Update

Using the above relations, I was able to develop the following expression for $I$:

$$begin{align*}I &= left(left[dfrac{a}{left(2pi r^2+a^2right)^frac{3}{2}} ast K_0left(bsqrt{2pi}rright)right] ast gleft(sqrt{2pi}rright)right) biggr{|}_{r=0}\
\
&= left(left[dfrac{a}{left(y^2+a^2right)^frac{3}{2}} ast K_0left(by right)right] ast gleft(yright)right) biggr{|}_{y=0}\
end{align*}$$

Where $*$ denotes 2D convolution of circularly symmetric functions.

The good news is that it is not necessary to figure out the Hankel Transform of $g(y)$. The bad news is that you have two 2D convolutions to compute. At least you only need the value at the origin, so maybe there is some savings there. You can also convert the 2D convolutions from polar coordinates back to cartesian coordinates, if you think there is some simplification there, but I suspect there is not.

$textbf {Edition of 05.12.2018}$

$$color{brown}{textbf{1. Integral representation of the Bessel function}}$$

Let us consider another way using the integral representation of the Bessel function $(69)$

$$J_0(z)=dfrac1piintlimits_0^pi cos(zsintheta),mathrm dtheta=dfrac2piintlimits_0^{pi/2} cos(zsintheta),mathrm dtheta.tag{1.1}$$

Substitution $sintheta = t$ transforms it to
$$J_0(z) = dfrac2piintlimits_0^1dfrac{cos(zt)}{sqrt{1-t^2}}mathrm dt.tag{1.2}$$
This leads to the formula
$$I = dfrac2piintlimits_0^1dfrac{C(t)}{sqrt{1-t^2}}dt,tag{1.3}$$
where
$$C(t) = intlimits_0^infty intlimits_0^infty f(x)g(y)xycos(xyt),mathrm{dx},mathrm{dy}.tag{1.4}$$
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, $$F(q)=mathcal H(f(x))= 2piintlimits_0^infty
xf(x)J_0(2pi qx),mathrm dx.tag{2.1}$$

Identity $(1.2)$ allows to write $(2.1)$ in the form of
$$F(q)= 4intlimits_0^1intlimits_0^infty dfrac{xf(x)cos(2pi qxt)}{sqrt{1-t^2}},mathrm dx,mathrm dt
= 4intlimits_0^1 dfrac{Phi(2pi qt)}{sqrt{1-t^2}}mathrm dt,tag{2.2}$$
where
$$Phi(p) = intlimits_0^infty xf(x)cos(px),mathrm dx.tag{2.3}$$
Let
$$f(x) = dfrac {e^{-sx}}x,$$
then
$$F(q)=mathcal Hleft(dfrac {e^{-sx}}xright) = 2pi intlimits_0^infty e^{-sx}J_0(2pi qx)mathrm dx = 2pimathcal L(J_0(2pi qx))
=dfrac{2pi}{sqrt{s^2+4pi^2q^2}}.tag{2.4}$$
On the other hand,
$$Phi(p) = intlimits_0^infty e^{-sx}cos(px),mathrm dx = mathcal L(cos(px)) = dfrac s{s^2+p^2}tag{2.5}$$
and
$$F(q) = 4intlimits_0^1 dfrac{Phi(2pi qt)}{sqrt{1-t^2}}mathrm dt
= 4intlimits_0^1 dfrac{s}{s^2+4pi^2q^2t^2}dfrac{mathrm dt}{sqrt{1-t^2}} = dfrac{2pi}{sqrt{s^2+4pi^2q^2}}tag{2.6}$$
(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 $f(x),$
$$intlimits_0^inftydfrac{xe^{-kx}}{x^2+b^2},mathrm dx = intlimits_0^inftydfrac{dfrac xbe^{-kbfrac xb}}{left(dfrac xbright)^2+1},mathrm ddfrac xb = intlimits_0^inftydfrac{xe^{-kbx}}{x^2+1},mathrm dx = F_1(kb),tag{3.1}$$
where
$$F_1(s)=left(dfracpi2 - mathrm{Si}(s)right)sin(s) - dfrac12mathrm{Ci}(s)cos(s))tag{3.2}$$
(see also Wolfram Alpha).

Then
$$intlimits_0^inftydfrac{xe^{-ax}cos(ytx)}{x^2+b^2},mathrm dx = dfrac{F_1(-ab+iybt)+F_1(-ab-iybt)}2,$$
$$C(t) = intlimits_0^infty yg(y)dfrac{F_1(-ab+iybt)+F_2(-ab-iybt)}2,mathrm{dy}.tag{3.3}$$
And the next step looks too complex.

$$color{brown}{textbf{4. Integration via the given function $g(y)$}}$$

For the given function $g(y),$
$$intlimits_0^infty yg(y)cos(xty),mathrm dy = intlimits_0^infty ye^{-y} L_l left( frac{2L+1}{l+L+1} y right) L_L left( frac{2l+1}{l+L+1} y right)cos(xty),mathrm dy = dfrac{G(1-ixt)+G(1+ixt)}2,tag{4.1}$$
where
$$G(s) = intlimits_0^infty ye^{-sy} L_l(y+py) L_L(y-py),mathrm dy,tag{4.2}$$
or
$$G(s) = -dfrac{mathrm d}{mathrm ds}mathcal Lbigl(L_l(y+py) L_L(y-py)bigr),tag{4.3}$$
where
$$p = frac{L-l}{l+L+1}tag{4.4},$$

Is known formula for the convolution
$$intlimits_0^y L_m(t)L_n(y-t),mathrm
dt = intlimits_0^y L_{m+n}(t),mathrm dt =
L_{m+n}(y)-L_{m+n+1}(y)tag{4.5}$$
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
$$C(t) = dfrac12 intlimits_0^infty xf(x)left(dfrac{P(1-ixt)}{(1-ixt)^{L+l+1}} + dfrac{P(1+ixt)}{(1+ixt)^{L+l+1}}right),mathrm dx,tag{4.7}$$
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
$$intlimits_0^inftydfrac{e^{-kx}}{x^2+b^2},mathrm dx = dfrac1b intlimits_0^inftydfrac{e^{-kbfrac xb}}{left(dfrac xbright)^2+1},mathrm ddfrac xb = dfrac1b intlimits_0^inftydfrac{e^{-kbx}}{x^2+1},mathrm dx = dfrac1b F_0(kb),tag{4.8}$$
where
$$F_0(s)=mathrm{Ci}(s)sin(s) + left(dfracpi2-mathrm{Si(s)}right)cos(s)tag{4.9}$$
(see also Wolfram Alpha),
$$intlimits_0^inftydfrac{e^{-kx}}{1pm ixt} = dfrac 1t intlimits_0^inftydfrac{e^{-frac kt xt}}{1pm ixt},mathrm d(xt) = dfrac 1t intlimits_0^inftydfrac{e^{-frac kt x}}{1pm ix},mathrm dx = dfrac 1t intlimits_0^inftydfrac{1mp ix}{x^2+1}e^{-frac kt x},mathrm dx,$$
$$intlimits_0^inftydfrac{e^{-kx}}{1pm ixt} = dfrac1t F_0left(dfrac ktright) mp dfrac it F_1left(dfrac ktright),tag{4.10}$$
$$H_m(s)=intlimits_0^inftydfrac{e^{-sx}}{(1pm ix)^m},mathrm dx
= -dfrac1{m-1} intlimits_0^infty e^{-sx},mathrm d(1pm ix)^{-m+1}$$
$$= -dfrac1{m-1} e^{-sx}(1pm ix)^{-m+1}bigg|_0^infty + dfrac s{m-1}intlimits_0^inftydfrac{e^{-sx}}{(1pm ix)^{m-1}},mathrm dx
= dfrac 1{m-1}left(1 + sH_{m-1}(s)right),$$
$$H_m(s)=sumlimits_{n=0}^{m-2}dfrac{(m-n-2)!}{(m-1)!}s^n+dfrac1{(m-1)!}(F_0(s)mp F_1(s)),tag{4.11}$$
$$intlimits_0^inftydfrac{e^{-kx}}{(1pm ixt)^m}dx = dfrac1t H_mleft(dfrac ktright).tag{4.12}$$

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.