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Mathematica tells me

$\int_0^{2\pi}e^{-ikr(\cos\theta_k\cos\theta+\sin\theta_k\sin\theta)}d\theta=2\pi J_0(kr)\ ,$

where $0<k\in\mathcal{R}$, $0\leq r\in\mathcal{R}$, $\theta_k\in\mathcal{R}$ and $J_0(x)$ is the $0$'th Bessel function.

I've been trying to prove this for ages without any luck. So far I tried to find a way to use the integral representations

$J_n(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{i(n\tau-x\sin\tau)}d\tau\ \ \ \ \ \ \ \ \ $ from Wikipedia

and

$J_0(z)=\frac{1}{\pi}\int_0^{\pi}\cos(z\sin\theta)d\theta=\frac{1}{\pi}\int_0^{\pi}\sin(z\sin\theta)d\theta\ \ \ \ \ \ \ \ \ $ from here,

but I can't quite get there. Can anyone point me in the right direction?

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$$\int_0^{2\pi} e^{-ikr(\cos\theta_k \cos\theta+\sin\theta_k\sin\theta)}d\theta = \int_0^{2\pi} e^{-ikr\cos(\theta-\theta_k)}d\theta = \int_0^{2\pi} e^{-ikr\cos\theta}d\theta= \int_0^{2\pi} [\cos(kr\cos\theta)-i\sin(kr\cos\theta)]d\theta =\int_{-\pi}^\pi [\cos(kr\cos\theta-i\sin(kr\cos\theta)]d\theta $$ Because the cosine is an even function of $\theta$ and therefore $\sin(..\cos\theta)$ an odd function of $\theta$, the imaginary part vanishes. $$\int_{-\pi}^{\pi}\cos(kr\cos\theta)d\theta =\int_0^{2\pi}\cos(kr\cos\theta)d\theta =\int_0^{2\pi}\cos(kr\sin\theta)d\theta $$ $$ =2\int_0^\pi\cos(kr\sin\theta)d\theta $$ as desired.

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