The Jacobi-Anger expansion is represented by
$$ e^{ix \cos\theta} = \sum_{n=-\infty}^{\infty} i^{n}J_{n}e^{in \theta}, $$
where
$$ J_n(x) = \frac{1}{2\pi} \int_{0}^{2\pi} \cos(x \sin\theta - n\theta)\ d\theta. $$
I was able to prove the identity above up to the following point:
Since $ e^{ix \cos\theta}$ is periodic, it can be rewritten as a Fourier series,
$$ e^{ix \cos\theta} = \sum_{n=-\infty}^{\infty} c_{n}e^{in \theta}, $$
and
$$ c_n(x) = \frac{1}{2\pi} \int_{0}^{2\pi} e^{ix \cos\theta}e^{-in \theta} \,d\theta. $$
So,
$$ c_n(x) = \frac{1}{2\pi} \int_{0}^{2\pi} e^{i(x \cos\theta -n \theta)} \,d\theta $$$$ = \frac{1}{2\pi} \int_{0}^{2\pi} \cos(x\cos\theta - n\theta)\,d\theta + i\sin(x\cos\theta - n\theta)\,d\theta. $$
Essentially, I want $c_n(x) = i^nJ_n(x)$, which will then prove that
$$ e^{ix \cos\theta} = \sum_{n=-\infty}^{\infty} i^{n}J_{n}e^{in \theta}, $$
however I'm not sure what the next step would be in proving this equality.
$\endgroup$1 Answer
$\begingroup$Looking at the integral representation of $J_n(x)$ you have, it's more natural to consider $$e^{ix\sin\theta}=\sum_{n=-\infty}^\infty a_n e^{in\theta},\qquad a_n=\frac{1}{2\pi}\int_{-\pi}^\pi e^{i(x\sin\theta-n\theta)}\,d\theta.$$ Clearly $\Im a_n=0$ (since $\theta\mapsto\sin(x\sin\theta-n\theta)$ is odd), so that we have $a_n=J_n(x)$ and $$e^{ix\cos\theta}=e^{ix\sin(\theta+\pi/2)}=\sum_{n=-\infty}^\infty a_n e^{in(\theta+\pi/2)}=\sum_{n=-\infty}^\infty i^n a_n e^{in\theta}.$$
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