Computation of some Fourier/Hankel transforms

Question

I need to compute the Fourier transform of the following functions :

$$f(x_1,x_2) = \frac{1}{a + (1-|x_1|^2-|x_2|^2)^2}$$

where $a>0$ is a positive constant. I have seen that because this function is radially symmetric, one way to compute its Fourier transform is to compute the Hankel 0-th order of the function

$$g(r) = \frac{1}{a + (1-r^2)^2}$$

and then $\mathcal{F}(f) = 2\mathbb{H}_0(g)$.

There is book called ''Tables of Integral Transforms'' that treats some Hankel transforms (chapter 8) but I was unable to find a satisfying formula for my specific problem. My intuition is to maybe decompose the rational fraction into more simple fractions and then use a known formula.

Answer 1

Following K.defaoite ideas, here is what I came up with :

$$g(r) = \frac{1}{2i\sqrt{a}}\bigg(\frac{1}{r^2-1 - i\sqrt{a}} - \frac{1}{r^2-1 + i\sqrt{a}}\bigg).$$

Now can find $b \in \mathbb{C}$ such that $b^2 = -1-i\sqrt{a}$ and so can construct $h$ as $$h(r) = 2i\sqrt{a}g(r) = \frac{1}{r^2+b^2} - \frac{1}{r^2+ (\overline{b})^2}$$

and so using the formula given by K.defaoite http://users.dimi.uniud.it/~giacomo.dellariccia/Table%20of%20contents/Piessens2000.pdf (can be found on the wikipedia page as well) we obtain that

$$\mathbb{H}_0(h)(x) = K_0(bx) - K_0(\overline{b}x).$$

But now, is there a way to explicitely compute this quantity ? I do not know enough about Bessel functions identities.

Answer 2

Here is the closest thing I could find.

There is a table of Hankel transforms here and they state that if $$f(r)=\frac{r^\nu}{(r^2+a^2)^{\mu+1}}$$ That its $\nu$th order Hankel transform is $$\mathcal{H}_\nu(f)(s)= \frac{s^\mu a^{\nu-\mu}}{2^\mu\Gamma(\mu+1)}K_{\nu-\mu}(as)$$

So i.e in the case $\nu=0,\mu=1,a=i$ you get $$f(r)=\frac{1}{(1-r^2)^2}$$ So then $$\mathcal{H}_0(f)(s)=\frac{-is}{2}K_{-1}\left(is\right)$$ And it is known $$K_{-1}(is)=\frac{\pi}{2}\left(J_1(s)+iY_1(s)\right)$$

Perhaps try completing the square in the denominator? I don't know what else to tell you.