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?
Best Answer
Using
$$\delta(f(x)) = \sum_{f(x_i)=0}\frac{1}{|f'(x_i)|}\delta(x-x_i)$$
we get that
$$\mathcal{F}\Bigr\{J_0(x)\Bigr\} = \frac{1}{2\pi}\left(\frac{1}{\left|\sin\Bigr(\cos^{-1}(k)\Bigr)\right|}+\frac{1}{\left|\sin\Bigr(2\pi-\cos^{-1}(k)\Bigr)\right|}\right) = \frac{1}{\pi}\frac{1}{\sqrt{1-k^2}}$$