Fourier series of a simple function

Question

Let $f : [- \pi , \pi) \to \mathbb{R}$ be the function $f = {\chi}_{(0 , \pi)} - {\chi}_{[- \pi , 0]}$ and let's extend $f$ to $\mathbb{R}$ such that $f$ is a $2 \pi$-periodic function. Then $f$ is an odd fuction for almost point and, calling $S f$ the Fourier series of $f$, we have that $$ S f(x) = \sum_{k = 1}^{\infty} b_k \sin k x $$ for all $x \in \mathbb{R}$, where $$ b_k = \frac{1}{\pi} \int_{[- \pi , \pi]} f(t) \sin k t \, dt $$ for all $k = 1 , 2 , \ldots$. So is not difficult to say that $$ S f(x) = \frac{4}{\pi} \sum_{k = 1}^{\infty} \frac{\sin (2 k - 1) x}{2 k - 1} $$ for all $x \in \mathbb{R}$. Obviously, $f$ is derivable on $\mathbb{R} \setminus \pi \mathbb{Z}$, being $$ \pi \mathbb{Z} = \{\pi k : k \in \mathbb{Z}\}\mbox{,} $$ so $S f = f$ on $\mathbb{R} \setminus \pi \mathbb{Z}$. But what happen in the points of the form $\pi k$, $k \in \mathbb{Z}$? Thank you very much.

Answer 1

In general, when you have a jump discontinuity, the Fourier series converges to the midpoint of this jump.

In your case, it would be $0$