Fourier transform of Bessel functions

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

Although this question was asked and answered quite a while ago, I thought that it might be useful to see an alternative development, one that uses classical analysis only and forgoes the use of Generalized Functions.

Here, we will find the Fourier Transform of $J_0(x)$ by first finding the Fourier Transform representation of $J_0(x)$ and subsequently invoking the Fourier Inversion Theorem.

We begin, as @Sasha began, with the integral representation

$$\begin{align} J_0(x)&=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{ix\sin \phi}\,d\phi\\\\ &=\frac{1}{\pi}\int_{0}^{\pi}\cos (x\sin \phi)\,d\phi \tag 1 \end{align}$$

where we exploited the fact that real part of the integrand is an even function of $k$ while the imaginary part is odd. Making the substitutions

$$\phi= \begin{cases} \arcsin (k),&\text{for}\,\,0\le\phi\le\pi/2\\\\ \pi-\arcsin (k),&\text{for}\,\,\pi/2\le\phi\le\pi \end{cases}$$

into $(1)$ yields

$$\begin{align} J_0(x)&=\frac{2}{\pi}\int_{0}^{1}\frac{1}{\sqrt{1-k^2}}\cos (kx)\,dk \tag 2\\\\ &=\frac{1}{\pi}\int_{-1}^{1}\frac{1}{\sqrt{1-k^2}}\cos (kx)\,dk \tag 3\\\\ &=\frac{1}{\pi}\int_{-1}^{1}\frac{1}{\sqrt{1-k^2}}e^{ikx}\,dk \tag 4\\\\ &=\frac{1}{2\pi}\int_{-1}^{1}\frac{2}{\sqrt{1-k^2}}e^{ikx}\,dk \tag 5\\\\ &=\frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\text{rect}\left(\frac{k}{2}\right)\frac{2}{\sqrt{1-k^2}}\right)e^{ikx}\,dk \tag 6\\\\ \end{align}$$

where $\text{rect}$ is the Rectangular Function. Finally, using the Fourier Inversion Theorem, we have

$$\bbox[5px,border:2px solid #C0A000]{\mathscr{F}\left(J_0(x)\right)(k)=\text{rect}\left(\frac{k}{2}\right)\frac{2}{\sqrt{1-k^2}}}$$

recovering the well-known result.

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NOTES:

In arriving at $(2)$, we wrote $\int_{0}^{\pi}\cos (x\sin(\phi))\,d\phi=\int_{0}^{\pi/2}\cos (x\sin(\phi))\,d\phi+\int_{\pi/2}^{\pi}\cos (x\sin(\phi))\,d\phi$, enforced the substitutions, and combined the resulting integrals.

In going from $(2)$ to $(3)$, we exploited the even property of the cosine.

In going from $(3)$ to $(4)$, we exploited the odd property of the sine.

In going from $(4)$ to $(5)$, we placed a factor of $1/2$ outside the integral and a factor of $2$ in the integrand.

In going from $(5)$ to $(6)$, we multiplied the integrand by the rectangle function and extended the limits.