This is a question that was [already asked on this site][1] but got no satisfactory answer. I would like to rephrase and show my own attempt. So the point is, consider the sequence
$a_k = \begin{cases} 1/k, & k \geq 1 \\ 0, & k \leq 0 \end{cases}.$
We want to show there is no Riemann-integrable function with this sequence as Fourier coefficients. I believe the way is to show that such a function $f$ cannot be bounded. In the first place, one can show that the convolution with the Fejer kernel of such a function diverges at 0. In fact, we have
$$\frac{1}{N+1}\sum_{k=1}^N\sum_{n=0}^N \widehat{f}(n)=\sum_{k=1}^N\frac{1}{k}-\frac{1}{N+1}\to\infty.$$
That convolution does not necessarily converge at $f(0)$, but it must converge to $f(x)$ for every $x$ such that $f$ is continuous at $x$. As $f$ is Riemann-integrable, we know it is continuous almost everywhere. Now, I would like to show that, getting sufficiently close to $0$, $\lim_{N\to\infty}f*F_N(x)$ gets arbitrarily big. Then $f$ would not be bounded. However, I do not know how to do this (or even if it is true), as the convergence of $f*F_N$ is not uniform. [1]: For a given sequence $(a_k)$, there is no Riemann integrable function f such that $\hat{f}(k) = a_k \forall k$