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.

proofof what seems like a non-trivial identity. Those voting to close because the question is "too trivial" are invited to make that triviality manifest by at least pointing to a semblance of an answer. $\endgroup$ – Igor Khavkine Aug 2 '15 at 18:10