What is the Fourier series of $\sin(1/x)$ in $[-\pi,\pi]$?

Question

What is the Fourier series of $\sin(1/x)$ (or $x^k\sin(1/x)$, where $k$ is a postive integer) in $[-\pi,\pi]$? This function evidently does not satisfy Dirichlet's conditions. However, Dirichlet's conditions are known to be sufficient but not necessary for the convergence of Fourier series. Is there any book (not too sophisticated) where the convergence of the Fourier series of these functions is mentioned or the series is given explicitly? The only reference I have found where the expansibility of $\sin(1/x)$ in Fourier series is mentioned is "Mathematical methods in the physical sciences" by M. L Boas (2005) (Chapter 7, page 358). Thanks in advance!

I had asked this question on MathOverflow (by mistake!) but somebody answered there saying there are no closed form expression for the series. But I would also like to know if there is any book (at a similar level as that of Boas) where the expansibility of these functions in Fourier series is mentioned. The book on Differential equations by George F. Simmons mentions these functions and how they do not satisfy Dirichlet's conditions but stops short of saying whether they still admit a Fourier series expansion.

Answer 1

Let's stick with the Riemann integral for the Fourier series coefficients. The function $\sin(1/x)$ can be assigned any value that you like at $x=0$, and that value does not matter because it will not affect the Riemann integral. After assigning some value to the function at $x=0$, the function is Riemann integrable on $[-\pi,\pi]$ because the set of discontinuities of $\sin(1/x)$ is of measure $0$. So there's no problem computing the Fourier series coefficients for this function on $[-\pi,\pi]$.

The Fourier series will converge to $\sin(1/x)$ pointwise everywhere on the set $S=[-\pi,0)\cup(0,\pi]$. In fact, it converges uniformly on every compact subset of $S$. These are all classical results of Fourier Analysis that require Riemann integrals only because the function is of bounded variation on every closed interval on $S$, or you can use differentiability of your function to assert convergence on such an interval. The simplest proof involves integration by parts along with boundedness near $x=0$ in order to establish convergence to the original function, ignoring $x=0$.