How to calculate the Fourier transform of the Kaiser-Bessel window?

Using a parity property, the Fourier integral can be written as \begin{equation} K=\frac{2}{I_0(\pi\alpha)}\int_0^{L/2}I_0\left(\pi \alpha \sqrt{1-{(2x/L)}^2}\right)\cos(2\pi fx)\,dx \end{equation} We use the quoted series expansion for the modified Bessel function to obtain after swaping integration and summation \begin{align} K&=\frac{2}{I_0(\pi\alpha)}\sum_{m=0}^\infty \frac{1}{(m!)^2}\left( \frac{\pi \alpha}{2} \right)^{2m}\int_0^{L/2}\left( 1-(2x/L)^2 \right)^{m}\cos(2\pi fx)\,dx\\ &=\frac{L}{I_0(\pi\alpha)}\sum_{m=0}^\infty \frac{1}{(m!)^2}\left( \frac{\pi \alpha}{2} \right)^{2m}\int_0^1\left( 1-t^2 \right)^{m}\cos(t\pi fL)\,dt \end{align} This cosine transform is tabulated in Ederlyi (TI 1.3.8) or can be related to an integral representation of the Bessel function: \begin{equation} J_{\nu}\left(z\right)=\frac{2(\tfrac{1}{2}z)^{\nu}}{\pi^{\frac{1}{2}}% \Gamma\left(\nu+\tfrac{1}{2}\right)}\int_{0}^{1}(1-t^{2})^{\nu-\frac{1}{2}}% \cos\left(zt\right)\mathrm{d}t \end{equation} With $\nu=m+1/2,z=\pi fL$ one obtains \begin{equation} \int_0^1\left( 1-t^2 \right)^{m}\cos(t\pi fL)\,dt=m!\sqrt{\pi}2^{-m+1/2}(\pi fL)^{-m-1/2}J_{m+1/2}(\pi f L) \end{equation} Then, after some simplifications, \begin{equation} K=\frac{L}{I_0(\pi\alpha)\sqrt{2fL}}\sum_{m=0}^\infty \frac{1}{m!}\left( \pi \alpha\right)^{2m}2^{-m}(\pi fL)^{-m}J_{m+1/2}(\pi f L) \end{equation} Such a series looks similar to the multiplication theorem for the Bessel functions: \begin{equation} J_{\nu}\left(\lambda Z\right)=\lambda^{\nu}\sum_{m=0}^{\infty}% \frac{(- 1)^{m}(\lambda^{2}-1)^{m}(\tfrac{1}{2}Z)^{m}}{m!}J_{\nu+ m}\left(Z\right) \end{equation} which is valid for any complex value of $\lambda$. We use $\nu=1/2,Z=\pi fL$ and $\lambda=\sqrt{1-\frac{\alpha^2}{f^2L^2}}$ with $\Im\lambda\ge0$ to write \begin{equation} K=\frac{L}{I_0(\pi\alpha)\sqrt{2fL}}\frac{1}{\left( 1-\frac{\alpha^2}{f^2L^2} \right)^{1/4}}J_{1/2}\left( \pi fL\sqrt{1-\frac{\alpha^2}{f^2L^2}} \right) \end{equation} and with the explicit expression for $J_{1/2}$, \begin{equation} K=\frac{L}{I_0(\pi\alpha)}\frac{\sin\left( \pi\sqrt{f^2L^2-\alpha^2} \right)}{\pi\sqrt{f^2L^2-\alpha^2} } \end{equation} which is valid for all the values of $f$. In particular, for $f<\alpha/L$, it is convenient to write the above expression as \begin{equation} K=\frac{L}{I_0(\pi\alpha)}\frac{\sinh\left( \pi\alpha\sqrt{1-f^2L^2/\alpha^2} \right)}{\pi\alpha\sqrt{1-f^2L^2/\alpha^2} } \end{equation} which is the proposed expression for the Fourier transform.