Why $\sum_{n\geq2}\frac{\cos nx}{\ln(n)}$ is a Fourier series for some Lebesgue integrable function?

Question

Why $\sum_{n\geq2}\frac{\cos nx}{\ln(n)}$ is a Fourier series of a Lebesgue integrable function? Does there exist any further information (like closed formula) of the corresponding function whose Fourier series would be $\sum_{n\geq2}\frac{\cos nx}{\ln(n)}$?

Answer 1

The series converges pointwise to an even function $f$ on $[-\pi,\pi]\setminus\{0\}$.

By the Dirichlet test, the series is uniformly convergent on any interval $[\delta,\pi]$ where $0 < \delta < \pi$. Furthermore, we have $f$ continuous on $[\delta,\pi]$. Thus, we can integrate termwise to obtain

$$ \int_\delta^\pi f(x) \, dx = \sum_{n=2}^\infty \frac{1}{\ln n}\int_\delta^\pi \cos nx \, dx = - \sum_{n=2}^\infty \frac{\sin n\delta}{n \ln n}$$

By a well-known theorem, the series on the RHS converges uniformly on $[0,\pi]$ since the coefficients $b_n = 1/(n \ln n)$ are monotonically decreasing and satisfy $nb_n \to 0$ as $n \to \infty$.

Therefore, we can interchange the limit as $\delta \to 0$ with the sum to obtain

$$\int_0^\pi f(x) \, dx = - \sum_{n=2}^\infty \lim_{\delta \to 0}\frac{\sin n\delta}{n \ln n}=0$$

This proves that $f$ is integrable on $[0,\pi]$ as well as $[-\pi,\pi]$ since it is even.

By a similar argument we can show that

$$\frac{2}{\pi}\int_0^\pi f(x) \cos nx \, dx = \frac{1}{\ln n}$$

Therefore, this is a Fourier series for $f$.