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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}

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I assume you mean $F = -1$ for $-\pi \le x \le 0$? – Eric Angle Dec 6 '12 at 16:25

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).

Mathematica graphics

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