Is there a way to get the Fourier transform / series of sinc(a*cos(t))? I'm trying to solve a partial differential equation via spectral methods, but it hinges on my ability to get an accurate Fourier series (up to arbitrary order) of a pretty messy function.
$$\mathrm{sinc}(a \cos(\theta)) = \frac{\sin\big({a \cos(\theta)}\big)}{a \cos(\theta)}$$
With $\mathrm{sinc}(0) = 1$
I've had some preliminary success using the Jacobi–Anger expansion for $\sin(a \cos(x))$
$$\mathrm{sinc}(a \cos(\theta)) = \frac{-2}{a} \sum_{n=1}^{\infty} (-1)^n \mathrm{J}_{2n-1}(a) \sec(\theta) \cos\big( (2n-1)\theta \big)$$
Now, my hope was to find the Fourier series expansion for $\sec(\theta) \cos\big((2n-1)\theta\big)$, and then rearrange the summations to keep the Fourier series and compute the infinite Jacobi-Anger series. But so far I haven't had much luck.
Any advice would be appreciated.
Best,
Kevin
 A: The function is even, so we only need the Fourier coefficients against the cosines.
$$\begin{split}
\int_{-\pi}^\pi e^{in\theta} \frac{\sin(a\cos\theta)}{\cos\theta}d\theta &= \int_{-\pi}^\pi e^{in\theta} \int_0^a\cos(u\cos\theta)du\,d\theta\\
&= \int_0^a\int_{-\pi}^\pi e^{in\theta}  \cos(u\cos\theta)d\theta\, du\\
&= \int_0^a \frac 1 2  \int_{-\pi}^\pi \left(e^{i(n\theta +u\cos\theta)} + e^{i(n\theta -u\cos\theta)}\right)d\theta\,du\\
&= \frac 1 2\int_0^a  \left( e^{in\frac \pi 2}\int_{-\pi}^{\pi}e^{-i(n\theta +u\sin\theta)}d\theta + e^{in\frac \pi 2}\int_{-\pi}^{\pi}e^{-i(n\theta -u\sin\theta)}d\theta\right)du\\
&= \frac 1 2 \int_0^a  2\pi i^n \left(J_{-n}(-u)+ J_{-n}(u)\right)du
\end{split}$$
Since $J_n$ is odd if $n$ is odd, and is even otherwise, the integral above vanishes if $n$ is odd. Thus let's assume that $n=2m$ is even. In that case,  using the fact that $J_{-n}(x)=(-1)^n J_n(x)$,
$$
\int_{-\pi}^\pi \cos(2m\theta) \frac{\sin(a\cos\theta)}{\cos\theta}d\theta =
2\pi (-1)^{m}\int_0^a  J_{2m}(u)du$$
We conclude that $$\boxed{\frac{\sin(a\cos\theta)}{a\cos\theta} = \frac 1 a\int_0^aJ_0(u)du +\frac 2 {a}\sum_{m\geq 1}(-1)^m\left(\int_0^aJ_{2m}(u)du\right)\cos(2m\theta)}$$
The two seem to numerically coincide.
Defining
$$
I_{2m}(a) = \int_0^a J_{2m}(u)\, \mathrm{d} u,
$$
we obtain that
$$
I_{2m}(a) = I_0(a) - 2 \sum_{k=1}^m J_{2k-1}(a)
$$
and
$$
I_0(a) = aJ_0(a) + \frac{\pi a}{2} [J_1(a)\mathbf{H}_0(a) - J_0(a)\mathbf{H}_1(a)],
$$
where $\mathbf{H}_0(a)$ and $\mathbf{H}_1(a)$ are Struve functions of order 0 and 1, respectively.
By doing so, we do not need to perform the integral numerically.
The derivations can be seen in Sec. 1.1.7 of W. Rosenheinrich, Tables of Some Indefinite Integral of Bessel Functions of Integer Order, (2017).
