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The solution to $\operatorname{sinc}(x)$$=a,0<a<\frac 2\pi$ involves inverting $ax-\sin(x)$ near $x=\frac\pi2 $ by transforming into $f_a(x)=a\left(x+\frac\pi2\right)-\sin \left(x+\frac\pi2\right)-\frac{\pi a}2+1=1+ax-\cos(x)$ with interval $L=f_a\left(\frac\pi2\right)=\frac{\pi a}2+1$. Define $h_a(f_a(x))=x$ on this interval with a complex Fourier series

$$h_a(x)=\frac{\pi^2 a+8}{16L}+\frac1{2L}\sum_{0\ne n\in\Bbb Z}e^\frac{\pi i n x}L\int_0^\frac\pi2te^{-\frac{\pi i n}Lf_a(t)}df_a(t)$$

integrating by parts:

$$h_a(x)=\frac{4\pi x+a\pi^2+8}{8(a\pi+2)}-\sum_{0\ne n\in\Bbb Z}\frac{ie^\frac{\pi i n (x-1)}L}{2\pi n}\int_0^\frac\pi2e^{-\frac{i\pi n}L(at-\cos(t)}dt=\frac{4\pi x+a\pi^2+8}{8(a\pi+2)}+\sum_{n=1}^\infty\frac1{\pi n}\int_0^\frac\pi2\sin\left(\frac{\pi n (x-1)}L-\frac{\pi n}L(a t-\cos(t)\right)dt$$

Here is a plot the complex Fourier series result that $f_a(h_a(x))=x$ and a numerical test for the second series inverting $\operatorname{sinc}(x)$ at a point. We find $\operatorname{sinc}^{-1}(x)=\frac\pi2+h_x\left(1-\frac{\pi x}2\right)$. However, no matter what is tried,

$$\int_0^\frac\pi2e^{i(ut+v\cos(t))}dt\text{ or }\int_0^\frac\pi2\sin(w+ut+v\cos(t))dt\tag 1$$

seemingly cannot be evaluated using Bessel, Struve, Anger Weber, and Lommel functions or otherwise as a closed form. For example,

$$\begin{align}\int_0^\frac\pi2 e^{i (u t-v\cos(t))}=\int_0^\frac\pi 2 \cos(ut)\cos(v\cos(t))-i\cos(ut)\sin(v\cos(t))+i\sin(ut)\cos(v\cos(t))+\sin(ut)\sin(v\cos(t))dt= \cos\left(\frac{\pi u}2\right) s_{0,u}(v)-u\sin\left(\frac{\pi u}2\right)s_{-1,u}(v)+\color{red}{\int_0^\frac\pi 2 \sin(u t)\sin(v\cos(t))dt}+i\color{red}{\int_0^\frac\pi 2\sin(ut)\cos(v\cos(t))dt}\end{align}$$

with Lommel $s_{a,b}(z)$ in Gradshteyn and Ryzhik 3.715,3.716 and can simplify into Anger Weber or Bessel functions. Additionally, a Jacobi Anger expansion uses the Bessel J function:

$$\int_0^\frac\pi2 e^{iv\cos(t)}e^{i ut}dt=-\sum_{m\in\Bbb Z}\frac{i^{m+1}}{m+u} J_m(v)\left(e^{\pi i \frac{m+u}2}-1\right)$$

but this is not a closed form for the Fourier series coefficients.

How can we evaluate $(1)$, the integrals in red, or the above series?

Тyma Gaidash
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    Hello Have you think about the Inverse laplace method see https://www.mdpi.com/2504-3110/5/2/43 ? Good luck in your investigation . – Barackouda Jun 09 '23 at 11:35
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    @ErikSatie That article is using the sinc function to approximate an inverse Laplace transform, but the question is about finding the Fourier coefficient “transform” of the sinc function. How would the article apply to the question? – Тyma Gaidash Jun 09 '23 at 11:53
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    Excellent question why not tackle the problem with https://en.wikipedia.org/wiki/Laplace_transform instead of Fourier series ? see https://math.stackexchange.com/questions/361229/compare-fourier-and-laplace-transform .See https://www.sciencedirect.com/science/article/abs/pii/S0377042723001796 – Barackouda Jun 09 '23 at 13:10

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