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In the text "Fourier Analysis: An Introduction" by Elias M. Stein and Rami Shakarchi. I'm having trouble attempting to verify the that $f(x)$ can be written as a Fourier Series in $\text{Proposition (1.2)}$ Note my initial approach to verify $\text{Proposition (1.2)}$ can be seen within, $\text{Lemma (1.3)}$

$\text{Proposition (1.2)}$:

Let $f(x) \, = X_{[a,b]}(x)$ be the Characteristic Function of the interval $[a,b] \subset [-\pi,\pi]$ this can be fully expressed as the following. $$ X_{[a,b]}(x) = \left\{ \begin{array}{lr} 1 & if\, \, \, \, \, \, x \in [a,b]\\ 0 & if \, \, otherwise \end{array} \right. $$

The Fourier series of $f$ is given in $(1.)$

$(1)$

$$ \, \, \, \, \, \, f(x)=X_{[a,b]} \sim \frac{b-a}{2 \pi} \, + \, \sum_{n \neq 0}^{\infty} \frac{e^{-ina}-e^{-inb}}{2 \pi in}e^{inx}.$$

$\text{Lemma (1.3)}$

$\text{Definition (2.1)}$

If the function $f$ is an integrable function on the interval $[-\pi,\pi]$, then the $n^{th}$ Fourier coefficient of $f$ can be written as: $$f(n)=a_n = \frac{1}{2 \pi}\int_{-\pi}^{\pi}f(\theta)e^{in\theta}d \theta.$$

$\text{Definition (2.2)}$

The Fourier series of function $f$ given in the case our integrable function lies on the interval $[-\pi,\pi]$ is: $$f(\theta)\sim \sum_{n=-\infty}^{\infty}a_ne^{in\theta}.$$

For integrating a product of two functions defined as the following for the case of definite integrals: $$\int_{a}^{b}u(x) \cdot v'(x)dx = (u(b) \cdot v(b) -u(a) \cdot v(a))-\int_{a}^{b}v(x) \cdot u'(x)dx.$$

By $\text{Definition (2.1)}$, our function $f(x)=X_{[a,b]}(x)$ can be written as the following Fourier coefficient: $$X_{[a,b]}(x)=\frac{1}{2 \pi}\int_{-\pi}^{\pi}f(\theta)e^{in\theta}d \theta.$$ Since our function $f$ now has a Fourier coefficient it can be written as the following Fourier Series: $$f(\theta) \sim \sum_{n=-\infty}^{\infty}((\frac{1}{2 \pi}\int_{-\pi}^{\pi}f(\theta)e^{in\theta}d \theta((e^{2 \pi inx/-\pi -pi)})).$$ Direct calculations on the integral by utilization of $IBP$ yield if $n \neq 0$: $$\frac{1}{2 \pi}\int_{-\pi}^{\pi}f(\theta)e^{in\theta}d \theta = \frac{1}{2 \pi}\Bigg[ \frac{\theta}{in}e^{in\theta}\Bigg]_{-\pi}^{\pi}+\frac{1}{2 \pi in} \int_{-\pi}^{\pi}e^{in\theta}d\theta=\frac{(-1)^{n+1}}{in}.$$

and since $n \neq 0$ our Fourier Coefficient becomes: $$f(0)=\frac{1}{2 \pi}\int_{-\pi}^{pi}\theta d\theta = 0.$$

Now our initial Fourier Series for our function $f$ is now: $$f(\theta) \sim \sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{in}e^{in\theta}e^{2\pi inx/b-a}).$$

My initial question specifically is how to show that the series in $(3.1)$ can be written as the series in $(3.2)$

$(3.2)$ $$ \, \, \, f(x)=X_{[a,b]} \sim \frac{b-a}{2 \pi} \, + \, \sum_{n \neq 0}^{\infty} \frac{e^{-ina}-e^{-inb}}{2 \pi in}e^{inx}.$$

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    $\begingroup$ I just want to point out that you seem to be dividing by $0$ multiple times throughout the post. $\endgroup$ Mar 19, 2017 at 0:27
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    $\begingroup$ well, you need some theorems about uniform convergence and point-wise convergence related to Fourier series. There is nothing about that in your question. In any case change of book, by example in Analysis II of Amann and Escher all the basic theory of Fourier is perfectly explained in all the details. $\endgroup$
    – Masacroso
    Mar 19, 2017 at 0:35
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    $\begingroup$ Can you edit your question, I think you are forgetting "-" minus sign in the Fourier coefficients. You are using circular coordinate $\theta$ and the linear coordinate $x$ in a confusing way. Your question is not clear either. $\endgroup$ Mar 19, 2017 at 0:57
  • $\begingroup$ I clearly defined in the beginning of my post the proposition I was trying verify and at the end of this post boiled down the crux of my question. In the middle was my initial approach. The initial error with the variables have been fixed note I misapplied the definition for Fourier Series on the interval $[-\pi,\pi]$. $\endgroup$
    – Zophikel
    Mar 19, 2017 at 1:17
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    $\begingroup$ If a function $f$ has left- and right-hand limits at a point $x$, along with left- and right-hand derivatives at that point, then the Fourier series of $f$ converges to the mean of the left- and right-hand derivatives.Your function has those properties. $\endgroup$ Mar 20, 2017 at 9:09

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As I recall, Stein and Shakarchi use the notation

$f$ ~ $\sum_n a_n e^{inx}$ to denote that the series on the right is the Fourier series of the function on the left? This notation doesn't imply anything about equality or convergence of that series; it's a formal expression. So what you really want to do is calculate the Fourier coefficients for $\chi_{[a,b]}$.

$a_0 = \frac{1}{2\pi} \int \chi_{[a,b]} = \frac{1}{2\pi} \int_a^b 1 = \frac{b-a}{2\pi}$

For $n \neq 0$,

$a_n = \frac{1}{2\pi} \int_a^b e^{-inx} dx = \frac{i}{2n\pi}(e^{-inb}-e^{-ina})$, which is what you wanted to show.

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  • $\begingroup$ Ahhh thank you I didn't essentially realize that calculating the Fourier Coefficients for $X_[a,b]$ $\endgroup$
    – Zophikel
    Apr 2, 2017 at 19:41

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