Show that $$\sum_\limits{n\ge 2}\frac{1}{\log n}\sin nx$$ is not the Fourier series of a Riemann integrable function.
I think this comes from the low increasing speed of $\log$ function. Analogously, $$\sum_\limits{n\ge 1}\frac{1}{n^\alpha}\sin nx$$ is not the Fourier series of a Riemann integrable function when $0<\alpha<1$.
And for the boundary situation,$$\sum_\limits{n\ge 1}\frac{1}{n}\sin nx$$
is a Fourier series of a Riemann integrable function, the sawtooth function.
Best Answer
1: Note that $\sum_\limits{n\ge 2}\frac{1}{\log n}\sin nx$ is not even the Fourier series of a Lebesgue integrable function, although it is convergent everywhere, uniformly convergent to a continuous function on any compact set avoiding integral multiples of $2\pi$ and its conjugate cosine series $\sum_\limits{n\ge 2}\frac{1}{\log n}\cos nx$ is a Fourier series of a Lebesgue integrable function which is $\infty$ at integral multiples of $2\pi$ and continuous everywhere else.
2: Both $S_\alpha(x) = \sum_\limits{n\ge 1}\frac{1}{n^{\alpha}}\sin nx$ and $C_\alpha(x) = \sum_\limits{n\ge 1}\frac{1}{n^{\alpha}}\cos nx$ are Fourier series of Lebesgue but not Riemann integrable functions for $0 < \alpha < 1$.
A direct proof for the second statement would go like this - using summation by parts we immediately see that both $S_\alpha(x), C_\alpha(x)$ converge uniformly on compact sets avoiding integral multiples of $2\pi$ to continuous periodic functions, while $S_\alpha(0) = 0, C_\alpha(0) = \infty$, so it is enough to show that near $0$, both $S_\alpha(x), C_\alpha(x)$ are $O(x^{\alpha -1})$ which shows that both are Lebesgue but not Riemann integrable, while similarly we show the partial sums are $O(x^{\alpha -1})$ so we can apply the Lebesgue dominated convergence theorem after integrating against $\sin(nx), \cos(nx)$ on say $[-\pi, \pi]$ and conclude that the Fourier coefficients are the ones appearing in the sum.
For the partial sums we separate them in sums up to $[\frac{1}{x}]$ and from $[\frac{1}{x}]$ to $N$ and we bound the first trivially ($|sin|, |cos| \leq 1$) and the second by partial summation using that $|\sum_A^B e^{inx}| = O(\frac{1}{x})$ if say $|x| < \frac{1}{2}, x \neq 0$, so we get an estimate $\sum_1^{[\frac{1}{x}]}\frac{1}{n^{\alpha}} + O((\frac{1}{x})(x^{\alpha})) = O_{\alpha}(x^{\alpha -1})$ with the implied constant depending only on ${\alpha}$ and then since we have an uniform $O$ (not depending on $N$) and the partial sums converge pointwise for each $x \neq 0$ we get the $O$ result for the limit too.
For both $S_\alpha(x), C_\alpha(x)$ we can actually get an asymptotic relation at $0$ with coefficients $\Gamma(1-\alpha)\sin\frac{\pi\alpha}{2}$ and $\Gamma(1-\alpha)\cos\frac{\pi\alpha}{2}$ (multiplying $x^{\alpha -1}$) using the Taylor expansion of $(1-re^{ix})^{\alpha - 1}, 0 < r < 1$ and noting that we can take a limit as $r$ goes to $1$ for any non-zero $x$ in $[-\pi, \pi]$ since the coefficients are positive decreasing to zero and a straightforward computation then gives the required asymptotic - this is enough to prove that the original sin series is not Riemann integrable.