My professor gives us the problem: Prove that $$ J_0(2\pi f_D\tau) \quad \underset{\mathcal{F}^{-1}}{\stackrel{\mathcal{F}}{\rightleftharpoons}} \quad \frac{1}{4\pi f_D\sqrt{1-(f/f_D)^2}}, $$ where $$ J_0(t) = \frac{1}{\pi} \int_0^\pi e^{jt\cos\theta} {~d\theta}. $$
My solving is:
Let us first arrange $J_0(t)$ into the form of $\displaystyle\int^\infty_{-\infty} f(\omega) e^{j\omega t} d\omega$. \begin{align} J_0(t) &= \frac{1}{\pi}\int_0^\pi e^{jt\cos\theta} {~d\theta} \label{1} \\ &= \frac{1}{\pi}\int_{-\frac\pi2}^{\frac\pi2} e^{jt\sin\phi} {~d\phi} \label{2} \\ &= \frac{1}{\pi}\int_{-\frac\pi2}^{\frac\pi2} \left[ \cos{\left(t\sin\phi\right)} + j\sin{\left(t\sin\phi\right)} \right] {~d\phi} \label{3} \\ &= \frac{1}{\pi}\int_{-\frac\pi2}^{\frac\pi2} \cos{\left(t\sin\phi\right)} {~d\phi} \label{4} \\ &= \frac{1}{\pi}\int_{-1}^{1} \frac{\cos{\left(t\omega\right)}}{\sqrt{1-\omega^2}} {~d\omega} \label{5} \\ &= \frac{1}{\pi}\int_{-1}^{1} \left[ \frac{1}{\sqrt{1-\omega^2}} \right] e^{j\omega t} {~d\omega} \label{6} \\ &= \frac{1}{\pi}\int_{-\infty}^{\infty} \left[ \Pi(\omega) \cdot \frac{1}{\sqrt{1-\omega^2}} \right] e^{j\omega t} {~d\omega} \label{7} \end{align} (1): from the zero-th order Bessel function of the 1st kind, (2): by the integration by substitution $\left(\cos\theta \to \sin\phi\right)$, (3): by Euler's formula, (4): by the fact that the imaginary part is odd, (5): by the integration by substitution $\left(\sin\phi \to \omega\right)$, (6): by the odd property of the sine function, and (7): by introducing the modified rect function $\Pi(x)$ whose value is $1$ for $x\in(-1,1)$ and $0$ for $x\in(-\infty,-1]\cup[1,\infty)$.
Substituting $t \to 2\pi f_D \tau$ in the eq. (\ref{7}), we have \begin{align} J_0(2{\pi}f_D{\tau}) &= \frac{1}{\pi}\int_{-\infty}^{\infty} \left[ \Pi(\omega) \cdot \frac{1}{\sqrt{1-\omega^2}} \right] e^{j\omega2{\pi}f_D{\tau}} {~d\omega} \label{8} \\ &= \frac{1}{{\pi}{f_D}}\int_{-\infty}^{\infty} \left[ \Pi\left(\frac{f}{f_D}\right) \cdot \frac{1}{\sqrt{1-(f/f_D)^2}} \right] e^{j2{\pi}f{\tau}} {~df} \label{9} \\ &= \frac{1}{{\pi}{f_D}} ~\mathcal{F}^{-1}\left\{ \Pi\left(\frac{f}{f_D}\right) \cdot \frac{1}{\sqrt{1-(f/f_D)^2}} \right\} \label{10} \end{align} (8): from the previous result, (9): by the integration by substitution $\left(\omega f_D \to f\right)$, and (10): by the definition of the inverse Fourier transform.
Since $J_0$ and its Fourier transform are absolutely integrable in the Lebesgue sense and it is continuous on the domain $(-f_D, f_D)$, \begin{align} \mathcal{F}\left\{ J_0(2{\pi}f_D{\tau}) \right\} &= \frac{1}{{\pi}{f_D}}\left\{ \Pi\left(\frac{f}{f_D}\right) \cdot \frac{1}{\sqrt{1-(f/f_D)^2}} \right\} \label{11} \\ &=\left\{ \begin{array}{cl} \displaystyle\frac{1}{{\pi}f_D\sqrt{1-(f/f_D)^2}} & \text{if }\left\vert f \right\vert < f_D\\ 0 & \text{otherwise} \end{array}\right. \label{12} \\ &= \mathbf{1}_{|f|<f_D} \cdot \frac{1}{{\pi}f_D\sqrt{1-(f/f_D)^2}} \label{13} \end{align} (11): by the Fourier inversion theorem, (12): by the definition of the function $\Pi$, and (13): by introducing the indicator function.
I really don't think where $4$ in the denominator is come from. Is my solving correct and the professor just made a mistake?
