Find the Fourier coefficients of $h(x) = \sum_{|n|\geq 2}^{\infty} \frac{\sin(nx)}{n\ln(n)}$

Question

Define h : $\Bbb R \to \Bbb R$ by $$h(x) = \sum_{|n|\geq 2}^{\infty} \frac{\sin(nx)}{n\ln(n)} \ \text{for all} \ x \in \Bbb R$$ The question asks to find the fourier coefficients of h

I have tried working it out the usual way. Given that $\hat h(n) = \frac{1}{2\pi} \int_{-\pi}^{\pi}h(t)e^{-int}dt$

It gives me $$ \hat h(n) = \frac{1}{2\pi}\int_{-\pi}^{\pi}\sum_{|n|\geq 2}^{\infty} \frac{\sin(nt)}{n\ln(n)}e^{-int}dt = \frac{1}{2\pi}\int_{-\pi}^{\pi}\sum_{|n|\geq 2}^{\infty}\frac{1}{2i} \frac{e^{int}-e^{-int}}{n\ln(n)}e^{-int}dt$$ $$ = \frac{1}{2\pi}\int_{-\pi}^{\pi}\sum_{|n|\geq 2}^{\infty}\frac{1}{2i} \frac{1-e^{-2int}}{n\ln(n)}dt = \frac{1}{2\pi}\int_{-\pi}^{\pi}\sum_{|n|\geq 2}^{\infty}\frac{1}{2i} \frac{1}{n\ln(n)}dt -\frac{1}{2\pi}\int_{-\pi}^{\pi}\sum_{|n|\geq 2}^{\infty}\frac{1}{2i} \frac{e^{-2int}}{n\ln(n)}dt $$

However, $$\sum_{n\geq 2}^{\infty}\frac{1}{n\ln(n)} = \infty$$

So I'm stuck here. How do I calculate the Fourier coefficients of h?

The first part I can still have an idea about the answer from just looking at the function. However the second part of the question asks to show that h $\notin C^1(\Bbb T)$. How to proceed with that?

Answer 1

The Fourier coefficients of the series are obviously:

$$\forall n\ge 0:\;a_n=0;\quad b_1=0,\;\forall n\ge 2:\;b_n=\frac1{n\ln n}.$$

$h(x)\ne C^1$ since $\lim_\limits{n\to\infty} n^2 |b_n|=\infty$.

Generally, $f(x)\in C^d$ iff $\lim_\limits{n\to\infty} n^{d+1} |\hat f(n)|<\infty$.

Answer 2

Your function $h$ is square integrable. If you express $\sin(nx)=\frac{e^{inx}-e^{-inx}}{2i}$ you get

$$ h(x)=\frac{1}{2i}\sum_{n\in\mathbb{Z}}\frac{e^{inx}}{n\log n}$$

from where you can easily extract the Fourier coefficients.