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I put the following integral into wolfram alpha to solve as a part of a larger project I'm working and got this very curious result

$ \int_{0}^{2\pi} \cos{(A \cos{(y - x)}}) dx = 2\pi J_0(A)$ for real A

Why is this true? Why does a Bessel function pop up here? The factor of $2\pi$ suggests some contour integration going on here but I'm having trouble seeing it

EDIT: missing a bracket should be cos(y-x) not cosy - x

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2 Answers 2

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I suspect a typo somewhere.

Let $$A \cos(y)-x=t\implies x=-t -A \cos(y)\implies dx=-dt$$ making $$\int \cos{(A \cos{y - x}})\, dx = -\int \cos(t) \,dt=\sin(t) +C$$ What is true is $$\int_{0}^{2\pi} \cos{(A \cos{y - x}})\, d\color{red}{y} = 2 \pi \cos (x) J_0(|A|)$$

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  • $\begingroup$ For real $A$, the absolute value is irrelevant because $J_0$ is an even function. For complex $A$, the use of absolute value seems incorrect. (Taking $a=i $ in WA gives an answer in terms of $I_0(1)$, not $J_0(1)$.) $\endgroup$ Jan 17, 2020 at 15:11
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For real values of the parameters the integrated function $$ f(x,y;A)=\cos{(A \cos{(x - y)}}) $$ is $\pi$ periodic with respect to $x$. Therefore: $$ \int_0^{2\pi}\cos(A \cos(x-y))dx=\int_{-y}^{-y+2\pi}\cos(A\cos t)dt =2\int_{0}^{\pi}\cos(A\sin t)dt=2\pi J_0(A), $$ the last equality following from the integral representation of the Bessel function.

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