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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

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    I looked through it but cannot filter that info to help me out with this proof. Do you mind using that pdf and placing it here and explaining it further to how it relates to this problem? Then I can ask some follow up question2012-10-26

1 Answers 1

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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} $

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    We actually need to invoke convergence in our proof in this problem. While your statement is true, it is missing in the proof2012-10-28