# Fourier Transform of Bessel Function

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?

• 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 $$\delta(f(x)) = \sum_{f(x_i)=0}\frac{1}{|f'(x_i)|}\delta(x-x_i)$$ – Ninad Munshi Sep 6 '20 at 7:38
• – Henry Feb 20 at 15:54

## 1 Answer

Using

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

we get that

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

• 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? – Joeseph123 Sep 6 '20 at 7:48
• @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 – Ninad Munshi Sep 6 '20 at 7:50
• @Joeseph123. A recent question: math.stackexchange.com/q/3814228/168433 – md2perpe Sep 6 '20 at 8:43