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Does anybody happen to know what the Fourier transform of $f(x) = \sin(x + \sin(x))$ is?

More generally, what is the Fourier transform of $g(x) = \sin(\phi_1 x + \mu \sin(\phi_2 x))$? ($\phi_1$, $\phi_2$ and $\mu$ are constants.)

I'm pretty sure I saw these written down somewhere, but Wolfram|Alpha insists that it cannot determine the result...

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up vote 5 down vote accepted

Since $\sin(x+\sin(x))$ is an odd function with period $2 \pi$, its Fourier series is $$ \sin(x+\sin(x)) = \sum_{n=1}^\infty c_n \sin(n x) $$ where $$ \begin{eqnarray} c_n &=& \frac{1}{\pi} \int_{-\pi}^\pi \sin(n x) \sin(x+\sin(x)) \mathrm{d} x = \frac{2}{\pi} \int_{0}^\pi \sin(n x) \sin(x+\sin(x)) \mathrm{d} x \\ &=& \frac{1}{\pi} \int_{0}^\pi \cos((n-1)x - \sin(x) ) \mathrm{d} x - \frac{1}{\pi} \int_{0}^\pi \cos((n+1)x + \sin(x) ) \mathrm{d} x \\ &=& J_{n-1}(1) - J_{n+1}(-1) = J_{n-1}(1) + (-1)^{n} J_{n+1}(1) \end{eqnarray} $$ where the last line was evaluated using Bessel integrals.

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I thought I remembered Bessel-J being in there somewhere... Is there an easy expression for the more complicated function? – MathematicalOrchid May 3 '12 at 8:54

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