I am given an integral representation of the Bessel Function $J_0$ as follows: $$J_0(x)=\frac{1}{2\pi}\int_0^{2\pi}{e^{ix\cos\theta}d\theta}$$

To compute the Fourier Transform, consider the integral: $$\mathscr F(J_0(x))=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{1}{2\pi}\int_0^{2\pi}{e^{ix\cos\theta}d\theta}\cdot e^{-ikx}dx$$ Combining the integrals and switching the order of integration: $$\mathscr F(J_0(x))=\frac{1}{4\pi^2}\int_0^{2\pi}\int_{-\infty}^{\infty}{e^{ix\cos\theta-ikx}dx}d\theta$$ We find that the inner integral is a Delta Function: $$\mathscr F(J_0(x))=\frac{1}{2\pi}\int_0^{2\pi}{\delta(\cos \theta-k)}d\theta$$ Applying a u-substitution $u=\cos \theta - k$, $u(0)=-k$, $u(2\pi)=-k$ and this is where I'm stuck: the limits of the integral are now equal. Have I made a mistake somewhere, or do I need to consider an infinitesimal interval around $-k$? If yes, how do I do this exactly?

  • 1
    $\begingroup$ Substitutions have to be broken up into branches where they are invertible. Cosine is not, on a $2\pi$ interval. But anyway, this is not the approach to take. Use $$\delta(f(x)) = \sum_{f(x_i)=0}\frac{1}{|f'(x_i)|}\delta(x-x_i)$$ $\endgroup$ Sep 6, 2020 at 7:38
  • $\begingroup$ Related: math.stackexchange.com/questions/78316/… $\endgroup$
    – Henry
    Feb 20, 2021 at 15:54
  • $\begingroup$ Does this answer your question? Fourier transform of Bessel functions $\endgroup$
    – Mark Viola
    Jul 4, 2021 at 15:41
  • $\begingroup$ For an alternative approach, SEE THIS. $\endgroup$
    – Mark Viola
    Jul 4, 2021 at 15:42

1 Answer 1



$$\delta(f(x)) = \sum_{f(x_i)=0}\frac{1}{|f'(x_i)|}\delta(x-x_i)$$

we get that

$$\mathcal{F}\Bigr\{J_0(x)\Bigr\} = \frac{1}{2\pi}\left(\frac{1}{\left|\sin\Bigr(\cos^{-1}(k)\Bigr)\right|}+\frac{1}{\left|\sin\Bigr(2\pi-\cos^{-1}(k)\Bigr)\right|}\right) = \frac{1}{\pi}\frac{1}{\sqrt{1-k^2}}$$

  • $\begingroup$ What is the name of this result: $\delta(f(x)) = \sum_{f(x_i)=0}\frac{1}{|f'(x_i)|}\delta(x-x_i)$ Could you give me a reference? $\endgroup$
    – Joeseph123
    Sep 6, 2020 at 7:48
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    $\begingroup$ @Joeseph123 it doesn't have a name as far as I know. You can "prove" it via u substitution (plus some series expansion) with the consideration for invertibility that I mentioned $\endgroup$ Sep 6, 2020 at 7:50
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    $\begingroup$ @Joeseph123. A recent question: math.stackexchange.com/q/3814228/168433 $\endgroup$
    – md2perpe
    Sep 6, 2020 at 8:43

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