pointwise convergence on Fourier series If $f(x)$ is a piecewise continous function in $[-l,l]$ how can we show that its indefinite integral $F(x) = \int _{-l}^x f(s) ds$ has a full Fourier series that converges pointwise?
And, how can we write this convergent series for $F(x)$ explicitly in terms of the Fourier coefficients $a_n, b_n$ of $f(x)$ if $a_0 = 0$?
Someting that came to mind is to applying a convergence theorem.
I really need help with the showing pointwise convergence. the proof below is good but is missing this and I cannot understand why we are dealing with pi instead of l in the proof below
 A: Theorem. Assume that $f$ is a piecewise continuous function on $[-\pi,\pi]$. For every $x_0$, $x\in [-\pi,\pi]$ there results
$$
\int_{x_0}^x f(t)\, dt = \int_{x_0}^x \frac{a_0}{2} + \sum_{n=1}^\infty \int_{x_0}^x \left( a_n \cos nt + b_n \sin nt \right)\, dt.
$$
Given $x_0$, the right-hand side converges uniformly in $[-\pi,\pi]$.
Proof. The function
$$
F(x)=\int_{x_0}^x \left( f(t)-\frac{a_0}{2} \right)\, dt, \quad x \in [-\pi,\pi]
$$
is continous, vanishes at $-\pi$ and at $\pi$, and we can extend it as a $2\pi$-periodic function. Moreover its derivative is continuous except at a finite number of points. By a well-know result, its Fourier series converges uniformly. If $A_n$ and $B_n$ are its Fourier coefficients, then 
$$
a_n = n B_n, \quad b_n = -n A_n.
$$
Hence
$$
\begin{align}
\int_{x_0}^x \left( f(t)- \frac{a_0}{2} \right)dt &= F(x)-F(x_0) =\\
&= \sum_{n=1}^\infty \left\{ A_n (\cos nx -\cos n x_0 ) + B_n (\sin nx - \sin n x_0 ) \right\}\\
&= \sum_{n=1}^\infty \left\{ -b_n \frac{\cos nx - \cos n x_0}{n}+a_n \frac{\sin nx - \sin nx_0}{n} \right\} \\
&=\sum_{n=1}^\infty \left\{ b_n \int_{x_0}^x \sin nt \, dt + a_n \int_{x_0}^x \cos nt \, dt \right\}.
\end{align}
$$
