Find an easy $f\in C(\mathbb{T})$ such that $\hat{f}\notin l^1$

Question

I'm looking for an easy example of a continuous function defined on the torus $\mathbb{T}$ such that its Fourier transform isn't in $l^1$. I'm aware of the fact that there are continuous functions defined on the torus $\mathbb{T}$ such that their Fourier transforms are not in $l^1$, e.g. the Lebesgue's counterexample (and, by the way, it is a consequence of the uniform boundedness principle). The point is that such a counterexample is constructed to obtain a continuous function such that its Fourier series diverges in a specific point. Now, if the Fourier trasform of $f$ is in $l^1$, total convergence of the Fourier series easily follows, so the uniformity of the convergence follows and so also the pointwise convergence follows. Then, it seems that the Lebesgue's counterexample actually provides much more of what I'm looking for. So I suspect that there must be a much easy counterexample for what I'm seeking. I strongly suspect that something like $$f:[-\pi,\pi]\rightarrow\mathbb{R}, t\mapsto\frac{1}{\log|\frac{t}{2\pi}|}$$ would do the job, but the calculation seems heavy... anyone can help me in proving or disproving such a fact or just to suggest an analogous simple example?

Answer 1

Let $f(x)=\sum_{n=2}^{\infty } \frac 1 {n \ln n} \sin nx$. Then the series converges uniformly by Theorem 7.2.2 in Fourier Series by Edwards. Hence $f$ is a continuous periodic function. Since $\sum_{n=2}^{\infty } \frac 1 {n \ln n}=\infty $ it follows that $\hat {f} \notin l^{1}$.