Proof of a Fourier pair with Bessel functions? How can we prove that the Fourier transform of the function
$$
f(x)
=
\begin{cases}
(a^2-x^2)^{c/2}  BesselJ[c,b\sqrt{a^2-x^2}] & \text{for }x^2 < a^2\\
0 & \text{otherwise}
\end{cases}
$$
is 
$$
\hat f(y)
=
\sqrt{2\pi a} (a^c) (b^c) (b^2+y^2)^{-c/2-1/4} BesselJ[c+1/2,a\sqrt{b^2+y^2}]
?
$$
This Fourier transform pair is given in the book
Formeln und Satze fur die speziellen Funktionen der mathematischer Physik
(Julius Springer, Berlin, 1943) p. 119. http://www.pokyrek.cz/Nijboer/Magnus_Hettinger.pdf
Numerical computation suggests this is correct.
I need this formula for $c=1$.
 A: The proof is not very complicated, but even a sketch needs more space than a comment. So here is a sketch.
I want to prove that
$$
\int_{-a}^{a} dx \ e^{i y x} \  f(x) = \hat{f}(y)
$$
with $f$ and $\hat{f}$ defined as in the question.
First one observes that it suffices to prove the equality for $a=1$. Then because of $f(-x)=f(x)$ and the symmetric integration interval one only has to prove that
$$
2 \int_{0}^{1} dx \ cos(y x) \  f(x) = \hat{f}(y).
$$
Expanding the Bessel function under the integral (use e.g. http://dlmf.nist.gov/10.2.E2) and exchanging sum and integration leaves us with integrals of the form ($m$ is the summation index)
$$
\int_{0}^{1} d x \ cos (y \ x) \ (1-x^2)^{c + m}
$$
which can be calculated by the so called Poisson's integral formula (found e.g. here:  http://dlmf.nist.gov/10.9.E4) resulting in essentially another Bessel function, $J_{c+m+1/2}(y)$.
We are thus confronted with a sum over Bessel functions each with argument $y$ attached with some factors.
After a (trivial) change of sign of these Bessel functions' argument (use http://dlmf.nist.gov/10.11.E1) we can evaluate the sum using the Multiplication Theorem for Bessel functions (see http://dlmf.nist.gov/10.23.E1), which finishes the proof.
