Discontinuities of $\sum_{n=2}^{\infty} \frac{\sin(nx)}{\ln(n)}$ Let us consider  the trigonometric series  $\phi(x)=\sum_{n=2}^{\infty} \frac{\sin(nx)}{\ln(n)}$. It is every where convergent. However it is not Lebesgue inntegrable in $[-\pi,\pi]$. 
What is the set of continuous points of $\phi$?
 A: On any interval $[\delta, \pi]$ with $\delta > 0$ we have
$$\left|\sum_{n=1}^m \sin nx\right| = \left|\frac{\sin \left(\frac{mx}{2} \right)\sin \left[\frac{(m+1)x}{2} \right]}{\sin \left(\frac{x}{2}\right)}\right| \leqslant \frac{1}{\sin \left(\frac{\delta}{2} \right)},$$
and the partial sums of $\sin nx$ are uniformly bounded.  Since $1/\ln(n)$ converges monotonically and uniformly to $0$, it follows by the Dirichlet test that the series converges uniformly.  A similar argument can be made for $x \in [-\pi,-\delta].$
Thus, $\phi$ is continuous at every point in $[-\pi,\pi] \setminus \{0\}$.
There must be a discontinuity at $x = 0$, since $\phi$ is not integrable on $[-\pi,\pi]$. 
Non-uniform convergence on $[-\pi,\pi]$
The convergence is not uniform on $[-\pi,\pi]$ -- otherwise we would have continuity at $x = 0$.  
This can be proved by showing the partial sums do not satisfy the Cauchy criterion uniformly if $x$ can be chosen arbitrarily close to $0$.  
For any $m \in \mathbb{N},$ let $x_m = \pi/(4m)$.  With $m \leqslant n \leqslant 2m$, we have $\pi/4 \leqslant nx_m \leqslant \pi/2$ and $1/ \sqrt{2} \leqslant \sin n x_m \leqslant 1$. 
Hence,
$$\left|\sum_{n = m}^{2m} \frac{\sin nx_m}{\ln(n)}\right| \geqslant \frac{1}{\sqrt{2}}\sum_{n=m}^{2m}\frac{1}{\ln(n)}$$
Since $\sum 1/\ln(n)$ diverges, the RHS cannot be arbitrarily small, regardless of the choice for $m.$ 
