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 $c=1$.

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    $\begingroup$ The question is asking for a proof of 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
  • $\begingroup$ Perhaps the OP may wish to share some initial attempts to resolve the problem? It looks likely that one can rescale one of $a$ or $b$ to equal $1$, although this only achieves a modest simplification. One approach would be to use the Bessel equation to work out the ODE that $f$ and the claimed value of $\hat f$ satisfy; if these Fourier transform to each other, and if one can verify suitable boundary conditions at infinity, one should be done. $\endgroup$ – Terry Tao Aug 3 '15 at 4:33
  • $\begingroup$ Another approach would be to multiply both $f$ and $\hat f$ by $t^c$ and sum over natural number $c$, using the generating function for the Bessel function; this should reduce matters (formally at least) to a simpler identity, at least for the case of natural number $c$ which is what the OP wants. $\endgroup$ – Terry Tao Aug 3 '15 at 4:38
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    $\begingroup$ Actually, I think the $c=1$ case might be obtainable by computing the Fourier transform of surface measure on the sphere $\{ (z,x) \in {\bf R}^4 \times {\bf R}: |z|^2 + x^2 = a^2 \}$ at $(b,0,0,0,y)$ in two ways: (i) by first taking Fourier transform in the z variable, and then in the x variable; (ii) by using spherical symmetry to replace $(b,0,0,0,y)$ with $(0,0,0,0,\sqrt{b^2+y^2})$ and then using cylindrical coordinates. $\endgroup$ – Terry Tao Aug 3 '15 at 4:49
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    $\begingroup$ The integral representation $J_\nu(z) = \frac{(z/2)^\nu}{\Gamma(\nu+\frac12)\sqrt{\pi}}\int_{-1}^1 e^{izt}(1-t^2)^{\nu-\frac12}dt$ may be useful it seems.... $\endgroup$ – Suvrit Aug 4 '15 at 21:37

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.

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    $\begingroup$ The series expansion approach sounds reasonable. There might yet be a slicker derivation using a clever choice of an integral representation. $\endgroup$ – Igor Khavkine Aug 10 '15 at 9:19
  • $\begingroup$ @IgorKhavkine I was inspired by the hints for proving Theorem 4.11.1 in "Special Functions" (1999) by Andrews, Askey, and Roy (p 217-218). $\endgroup$ – Johannes Trost Aug 10 '15 at 14:49
  • $\begingroup$ Many thanks to @Johannes for the advice how to prove the relation. It seems we do not need the "(trivial) change of sign". Also, I wanted to prove the Multiplication Theorem. This can be done by expanding the Bessel functions and comparing the coefficients at matching powers. Thanks a lot. $\endgroup$ – Pavel Aug 17 '15 at 14:33

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