How can I find Fourier series and convergence of function $f(x)= \min \{0, \cos x \}$ ? Because it is an even function I am expanding it only for cosine. Doing that I get $a_{0} = 0$, after that I did the following : $$b_n =\frac{1}{2\pi}\int\limits_0^\pi\ \cos x \cos nx\ dx = \int\limits_0^\pi\ \frac{1}{4\pi} (\cos(x(1+n )+\cos (x(1-n))\ dx = \frac{\sin(\pi(n+1))}{4\pi(n+1)} + \frac{\sin(\pi(n-1))}{4\pi(n-1)} $$
If this is correct I get a series that I can't find the convergence.
Note that $\cos$ is $2\pi$-periodic and that on $[0, 2\pi]$, $\min(0, \cos x) = 0$ if $x < \pi/2$ or $x > 3\pi/2$ and $\min(0, \cos x ) = \cos x$ otherwise.
So
$$b_n = \frac{1}{\pi} \int_0^{2\pi} \min(0, \cos x) \cos nx \, dx = \frac{1}{\pi} \int_{\pi / 2}^{3\pi / 2} \cos x \cos nx \, dx.$$