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.


2 Answers 2


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.


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.

  • $\begingroup$ Thank you! I receive your notification. I'm not able to use this site as much as I'd like these days unfortunately, but your answer is great. $\endgroup$
    – MattyZ
    Jul 17, 2015 at 1:24
  • 1
    $\begingroup$ You're welcome! My pleasure. It gives another way to see the transform. $\endgroup$
    – Mark Viola
    Jul 17, 2015 at 1:49
  • $\begingroup$ I was trying to compute the Fourier transform of the zeroth order Bessel function of second kind and stumbled on this. Your solution is really elegant and adaptable to my problem as well. Thank You! One small question, I don't see how your second substitution works for going from equation (1) to (2). So instead I chose $$\phi= \begin{cases} \arcsin (k),&\text{for}\,\,0\le\phi\le\pi/2\\\\ \pi/2+\arccos (k),&\text{for}\,\,\pi/2\le\phi\le\pi \end{cases}$$ Great approach nevertheless! $\endgroup$ Jun 30, 2017 at 2:11
  • $\begingroup$ @MandarKulkarni Thank you for the nice note. And you're welcome. Pleased to hear that this was useful. There was a typo in that substitution; I've edited. It should be $$\phi= \begin{cases} \arcsin (k),&\text{for}\,\,0\le\phi\le\pi/2\\\\ \pi-\arcsin (k),&\text{for}\,\,\pi/2\le\phi\le\pi \end{cases}$$ $\endgroup$
    – Mark Viola
    Jun 30, 2017 at 2:33
  • $\begingroup$ Yes the edited form works. $\endgroup$ Jun 30, 2017 at 2:38

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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    $\begingroup$ 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... $\endgroup$ Mar 12, 2014 at 4:45
  • $\begingroup$ @JeffFaraci From the convolution theorem, we know that the Fourier Transform of a product of functions is equal the convolution of the Fourier Transforms. Apply this first to find the Fourier Transform of $J_0^2$. Then apply it once more to find the Fourier Transform of $J_0^3$. $\endgroup$
    – Mark Viola
    Jun 27 at 13:13

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