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I'm curious as to how the Fourier transform of the various types of Bessel functions would be calculated. The Wikipedia page on the Fourier transform gives the transform of $J_o(x)$ as being $\frac{2rect(\pi\zeta)}{\sqrt{1-4\pi^2\zeta^2}}$. I've searched the web some but I seem to be unable to find a derivation for that value. Since $J_o$ only "damps out" to zero at infinity I'm guessing the transform must be computed using some kind of generalized function representation of the Bessel function. Any references would be much appreciated.

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Using integral representation: $$ J_0(x) = \frac{1}{2 \pi} \int_{-\pi}^\pi \mathrm{e}^{i x \sin \tau} \mathrm{d} \tau $$ Thus the Fourier transform: $$ \begin{eqnarray} \mathcal{F}_x(J_0(x))(\omega) &=& \int_{-\infty}^\infty J_0(x) \mathrm{e}^{i \omega x} \mathrm{d} x = \frac{1}{2 \pi} \int_{-\pi}^\pi \mathrm{d} \tau \, \mathcal{F}_x(\mathrm{e}^{i x \sin \tau})(\omega) = \frac{1}{2\pi} \int_{-\pi}^\pi \mathrm{d} \tau \, \left( 2 \pi \right) \delta\left( \omega + \sin(\tau) \right) \\ &=& \int_{-\pi}^\pi \mathbf{1}_{-1 \le \omega \le 1} \delta\left( \omega + \sin(\tau) \right) \,\, \mathrm{d} \tau \\ &=& \int_{-\pi}^\pi \mathbf{1}_{-1 \le \omega \le 1} \frac{1}{\vert \cos(\tau) \vert} \left( \delta\left( \arcsin \omega + \tau \right) + \delta\left( \arcsin \omega - \operatorname{sign}(\omega) \pi + \tau \right) \right)\,\, \mathrm{d} \tau \\ &=& \mathbf{1}_{-1 \le \omega \le 1} \frac{2}{\sqrt{1-\omega^2}} \end{eqnarray} $$

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Thanks, that helps! –  Bitrex Nov 2 '11 at 21:06
    
Hi, I am trying to do this problem, but instead I am trying $$ \int_{-\infty}^\infty J^3_0(x)e^{i\omega x} dx. $$ Any idea how I would go about this? THanks... –  Integrals Mar 12 at 4:45
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