Real Fourier series convergence I have problem with determining function that the Fourier series is convergent to. My function is $$f(x) = \frac{\cos(1/x)}{(\ln(|x|/4))^2} $$
for $x\in [-\pi, \pi)\setminus \{0\}$ and $f(0) = 0$. 
Using Jordan's criterion, I can show that $S_k f(x) \to f(x)$ for any $x\neq 0$. What about $x=0$? The oscilasions are so big there, that I can't say anything about that point. Will variation be bounded around that point? Maybe there's other way to do this?
 A: Two sufficient conditions for the convergence of a Fourier series at a point $x$ to the mean of the left- and right-hand limits $[f(x+) + f(x-)]/2$ are the Jordan criterion, that $f$ has bounded variation locally, and the Dini criterion, that $\phi(s)/s$ is absolutely integrable in a neighborhood of $0$ where
$$\phi(s) = f(x+s) + f(x-s) - [f(x+) + f(x-)]$$
We will pursue the Dini criterion here to show that the Fourier series is convergent at $x = 0$.
Note that $f$ is continuous at $x = 0$ with $f(0) = 0$.  Since $f$ is an even function we have $\phi(s) = 2f(s)$. The Dini criterion reduces to showing the integral $\int_0^\pi \frac{|\phi(s)|}{s} \, ds $ converges.  
Since $|\phi(s)|/s$ is continuous on $[\delta,\pi]$ for $0 < \delta < \pi < 4$, it is enough to show that  we have convergence of 
$$\tag{*} \int_0^\delta \frac{|\phi(s)|}{s} \, ds = \int_0^\delta \frac{2|\cos(1/s)|}{s(\ln(s/4))^2} \, ds = 2\int_{4/\delta}^\infty \frac{|\cos(u/4|}{u (\ln (1/u))^2} \, du$$
where the integral on the RHS was obtained by the variable transformation $u = 4/s$.
For $u > 4/\delta > 1$ we have 
$$\frac{|\cos(u/4)|}{u (\ln (1/u))^2} = \frac{|\cos(u/4)|}{u (\ln u)^2} \leqslant \frac{1}{u (\ln u)^2},$$
and
$$\int_{4/\delta}^\infty\frac{1}{u (\ln u)^2} = \frac{1}{\ln(4/\delta)}$$
By the comparison test, the integral (*) is convergent.
