Not Lebesgue integrable function

Let $x \in \mathbb{R}$. Given the series:

\begin{equation} \sum_{n=2}^{\infty} \frac{\sin nx}{\ln n} \end{equation}

1. Prove that $(1)$ converges forall $x \in \mathbb{R}$.
2. Prove that $(1)$ is not a Fourier series of a Lebesgue integrable function.

Solution

Well this is a pretty known exercise. 

1. Let $\kappa \in \mathbb{Z}$. We note that for $x=\kappa \pi$ the series trivially converges since $\sin \kappa \pi =0$. For all other $x$ we are using Dirichlet's test. It is well known that $\sum \limits_{m=1}^{n} \sin mx$ is bounded and we also note that $\frac{1}{\ln n} \searrow 0$. The result follows.


2.  Suppose that such $f$ exists. Integrating (we can do that since Fourier series is integrated term by term) we take a continuous and of bounded variance function. The Fourier series of this function should converge to $0$. This is not the case here since $\sum \limits_{n=2}^{\infty} \frac{1}{n \ln n}$ is known to diverge. Thus, such function does not exist.

Notes: 

1. The weakest result also holds. That is, there does not exist a Riemann integrable function such that this series is its Fourier series. This follows immediately from Parseval's identity.


2. We have seen a powerful theorem regarding uniform converge on a very similar series here . 
3. That the sum $\sum \limits_{m=1}^{n} \sin mx$ is bounded can be seen as follows:
   
   \begin{align*}
   \left | \sum_{m=1}^{n}  \sin mx\right | &=\left | \frac{\cos \frac{x}{2} - \cos \frac{(n+1)x}{2}}{2 \sin \frac{x}{2}} \right | \\
    &\leq \frac{1}{\sin \frac{a}{2}}
   \end{align*}
   
   A relative discussion can be found at mathimatikoi.org . (discussing the boundness of this particular sum)