For $x\in[-\pi,\pi]$,
$$F(x)=\left\{
\begin{array}{cl}
-1 & \text{for}~-\pi\leq x\leq 0\\
1 & \text{for}~0\leq x\leq \pi
\end{array}\right..$$
To what value does the Fourier series converge at the point $x=\pi$?
I have found the fourier series as show below and I think it converges to 0 on the conditions give but I'm not entirely sure.
$f(-x)=-f(x)$ so we have and odd function and expect $a_n=0$.
\begin{align}
b_n
&= \frac1\pi \left[\int^\pi_{-\pi}f(x)\sin(nx)dx\right] \\
&= \frac1\pi \left[\int^0_{-\pi}-\sin(nx)dx\right]
+ \frac1\pi \left[\int^\pi_0\sin(nx)dx\right] \\
&= -\frac1\pi \left[-\frac{\cos(nx)}{n}\right]^0_\pi
+ \frac1\pi \left[-\frac{\cos(nx)}{n}\right]^\pi_0 \\
&= \frac{1-(-1)^n}{n\pi} – \frac{(-1)^n-1}{n\pi} \\
&= \frac{1-(-1)^n}{2n\pi}.
\end{align}
$$b_{2m}=0,\quad b_{2m+1}=\frac{4}{(2m+1)\pi}$$
\begin{align}
f(x)
&=\frac4\pi\left(\sin x+\frac{1}{3}\sin(3x)+\frac{1}{5}\sin(5x)+\cdots\right)\\
&=\frac4\pi \sum^{\infty}_{m=0} \frac{1}{2m+1}\sin((2m+1)x)
\end{align}
Best Answer
A classical theorem on pointwise convergence of Fourier series says that if $f(x)$ is piecewise smooth on $(-\ell,\ell)$, then the Fourier series of $f$ converges pointwise on $(-\ell,\ell)$. Moreover, the value to which the Fourier series converges at $x=x_0$ is $${f(x_0^+)+f(x_0^-)\over 2},$$ where the superscripts denote the one-sided limits $$f(x_0^+):=\lim_{x\to x_0^+}f(x)\quad\text{ and }\quad f(x_0^-):=\lim_{x\to x_0^-}f(x).$$
In other words, if $x=x_0$ is a point of continuity of $f$, then its Fourier series converges to $f(x_0)$ there, but if $x=x_0$ is a point of (suitable type of) discontinuity of $f$, then its Fourier series converges to the average of the left- and right-and limits of $f$ at $x=x_0$.
Combining this with the fact that the Fourier series of $f$ on $(-\ell,\ell)$ corresponds to the periodic extension $f_\text{ext}$ of $f$ on $\mathbb{R}$, we see that at $x=\pi$, there is a jump discontinuity in $f_\text{ext}$ with $${f_\text{ext}(\pi^+)+f_\text{ext}(\pi^-)\over 2}=0.$$ Hence, the Fourier series of the given $f$ converges pointwise to zero at $x=\pi$. The graph below demonstrates this ($f_\text{ext}$ in black and a partial Fourier sum in blue).