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

According to Wikipedia: https://en.wikipedia.org/wiki/Kaiser_window, the Fourier transform of the Kaiser-Bessel window

$$w_0(x) := \left\{ \begin{array}{**lr**} \frac{I_0(\pi \alpha \sqrt{1-{(2x/L)}^2})}{I_0(\pi \alpha)}, & |x| \leq \frac{L}{2}\\ 0, & |x| > \frac{L}{2}\\ \end{array} \right.$$

is

$$W(f) := \frac{L\cdot \sinh(\pi\alpha\sqrt{1-(Lf/\alpha)^2})}{I_0(\pi\alpha) \cdot \pi\alpha\sqrt{1-(Lf/\alpha)^2}}$$,

where $$I_0(x) := \sum\limits_{m=0}^{\infty}\frac{1}{(m!)^2}(\frac{x}{2})^{2m}$$ is the zeroth-order modified Bessel function of the first kind (https://en.wikipedia.org/wiki/Bessel_function#Modified_Bessel_functions:_I%CE%B1,_K%CE%B1).

I have made many attempts but fail to deduce the closed-form solution of the Fourier transform of the Kaiser-Bessel window. First, I tried to expand $$I_0(x)$$, which was quite difficult to calculate since it involved infinite series. Second, I tried to calculate the Fourier transform from the opposite side, i.e., $$W(f)$$, but got no result. Finally, I tried to use Mathematica Professional 12, but it failed to offer a result after long time calculation.

I have also found some literature online, e.g.,

[1] On the use of the I0-sinh window for spectrum analysis, IEEE Transactions on Acoustics, Speech, and Signal Processing, J. Kaiser, R. Schafer, 1980, doi: 10.1109/TASSP.1980.1163349.

[2] On the use of windows for harmonic analysis with the discrete Fourier transform, Proceedings of the IEEE, F.J. Harris, 1978, doi: 10.1109/PROC.1978.10837.

[4] Verification of Fourier transformation of Io-sinh function (MathStackExchange; No answer yet).

Neither of them deduces the closed form of the Fourier transform of the Kaiser-Bessel window. Some references, e.g., [1] and [3], point to an old unavailable book named "System Analysis by Digital Computer". Kaiser-Bessel window is a widely used window function in signal processing. However, it is really difficult to find relevant literature online. Additionally, I am not sure whether the Fourier transform provided in https://en.wikipedia.org/wiki/Kaiser_window is a closed-form or an approximation. I deeply appreciate your help or hints.

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