TITLE: Fourier Series $\sin(\sin(x))$
QUESTION [12 upvotes]: Can anyone find the Fourier Series of $ \sin(\sin(x))$?
I have tried evaluating the integrals to determine the coefficients of each of the coefficients of the sine waves, but have no idea where to start computing the integrals.

REPLY [18 votes]: First note that $f(x)$ is anti-symmetric so the series only has $\sin$'s, and that only odd $n$'s will appear since $f(x-\pi)=-f(x)$.
The integrals give combinations of the Bessel J functions, at $x=1$. The coefficient of $\sin nx$ is:
\begin{align} 
b_n& = \frac{1}{\pi}\int_{-\pi}^{\pi} \sin(\sin(x))\sin(nx) \, dx \\
& = \frac{1}{\pi}\int_{-\pi}^{\pi} \cos(nx-\sin(x)) - \cos(nx)\cos(\sin(x))\, dx \\ 
& =2 J_n(1) 
\end{align}
Since:
$$J_n(1) = \frac{1}{\pi} \int_0^\pi \cos(nx - \sin x)\, dx\ ,$$
and the second integral becomes $0$ for odd $n$, as it is anti-symmetric around $\pm \pi/2$. 
Thus we have:
$$\large\color{blue}{\sin(\sin(x)) = 2\sum_{k=0}^\infty J_{2k+1}(1)\sin \left((2k+1)x\right)}$$
P.S: This gives an interesting identity for the sum of all odd $J_k(1)$:
$$\sin(1)=2\sum_{k=0}^\infty (-1)^kJ_{2k+1}(1)$$