As part of one problem I need to find the widest subset of $\mathbb{R}$ on which the obtained Fourier series can be integrated and derived term by term.
I found that it has something to do with uniform convergence, but I can't really apply anything I've read about it so far.
The Fourier series is: $$f(t)=\frac{2}{\pi}+\frac{4}{\pi} \sum_{n=1}^{+\infty}\left ( \frac{(-1)^{n-1}}{4n^2-1} \right )\cos(2nt)$$
Note: I can prove absolute convergence (from which uniform convergence follows) but I don't see how I can apply it to certain interval $[a,b]$ where I can discuss the bounds $a$ and $b$.
Notice that for all $t \in \Bbb R$,
$$\left| \left ( \frac{(-1)^{n-1}}{4n^2-1} \right )\cos(2nt) \right| \le \left| \frac{(-1)^{n-1}}{4n^2-1} \right| \underbrace {|\cos (2nt)|} _{\le 1} \le \left| \frac{(-1)^{n-1}}{4n^2-1} \right| = \frac 1 {4n^2-1}$$
and $\sum \frac 1 {4n^2-1}$ is convergent (use the limit comparison test and compare it with $\sum \frac 1 {n^2}$, which is known to be convergent), so by Weierstrass's M-test the series $\sum \limits _{n=1}^\infty \left ( \frac{(-1)^{n-1}}{4n^2-1} \right )\cos(2nt)$ converges absolutely and uniformly on $\Bbb R$, therefore it may be integrated term by term on any interval $[a,b] \subset \Bbb R$.