TITLE: The Fourier Transform of a Bessel function: $J_a(be^{-x}), a,b>0$
QUESTION [6 upvotes]: Currently we require the Fourier transform  $g(p)$ of  $f(x)=J_a(be^{-x}), a,b>0$
$$g(p)=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}e^{ipx} f(x) dx$$
We could not find it in Tables and Handbooks. Mathematica gives it in terms of Gamma functions and Gauss Hyper geometric function. A closer look suggested that Hypergeometric function was actually a hoax. As we were finally interested in $|g(p)|^2$, it was a pleasant surprise to get
$$|g(p)|^2=\frac{1}{2\pi(p^2+a^2)},~~~~~(1)$$
which is independent of $b$ !, even the verification of (1) by
Mathematica in analytic mode  for various sets of $a,b,p$ has been time consuming.
The question is how to prove (1) and get $g(p)$ by hand?
EDIT:
By Integrate of  Mathematica, I finally get
$$g(p)=\frac{1}{2\sqrt{2\pi}} \left(\frac{2}{b}\right)^{ip} \frac{\Gamma[(a+ip)/2]}{\Gamma[1+(a-ip)/2]}$$
EDIT:
Here is the Mathematica out put, see the pre=factor of hypergeometric function is 0.
g(p)=1/2 Gamma[
1/2 (a + I p)] ((2^(I p) (1/b^2)^((I p)/2))/
Gamma[1/2 (2 + a - I p)] +
2^-a (-(1/b^2)^(-a/2) + b^
a) HypergeometricPFQRegularized[{1/2 (a + I p)}, {1 + a,
1/2 (2 + a + I p)}, -(b^2/4)])

REPLY [2 votes]: The Fourier transform has a closed-form, but my derivation lacks rigor.
Eliminate the b-dependence by translation: $x \to x+\log{(1/b)}.$ Thus
$$ g(p):= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{ipx}J_a(be^{-x})dx=$$
$$=\frac{(\log{(1/b)})^{ip}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{ipx}J_a(e^{-x})dx = \frac{(\log{(1/b)})^{ip}}{\sqrt{2\pi}}G(a,p) $$
where $G(a,p)$ has been defined so that we can deal with a 2-variable integral now.
Use the known expansion (Gradshteyn & Rhyzik 8.411.10)
$$J_a(z) = \frac{(z/2)^a}{\Gamma(a+1/2)\Gamma(1/2)} \int_{-1}^1e^{izt}(1-t)^{a-1/2}dt$$
Insert this into $G(a,p).$ Interchange integrals.  Do the inner integral explicitly,
$$ \int_{-\infty}^\infty e^{ipx}e^{-ax}e^{ie^{-x}} = (-it)^{-a+ip}\Gamma(a-ip)$$
Here, though, the second non-rigorous step has been taken (the first is that the intergrals can be intergchanged).  Mathematica says the previous eq. is true for t with a non-zero imaginary part.  I'm assuming that someone good with contour integration can justify assuming completely real t is O.K.  Mathematica can do the remaining integral as well.  Collecting,
$$G(a,p):=\int_{-\infty}^\infty e^{ipx}J_a(e^{-x})dx=$$
$$=\frac{ 2^{-a} (-i)^{-a+ip}\ \Gamma(1/2+(ip-a)/2)}{2\Gamma(1/2)\Gamma(1+(ip+a)/2)}  (-1)^{-a}\big((-1)^{a}+e^{-p\pi})\big)\Gamma(a-ip).$$
(Hopefully there are no typos.)
As a numerical check, I let a=3/10 and p=6/10.  For the integral I chopped the integral limits at $\pm$ 1200, used 30 digits precision, and MinRecursion of 5 in Mathematica's NIntegrate.  I got 14 digits of agreement from the integral, and the explicit form presented.
Out of time; hope this helps.