Piecewise function into Fourier Series and evaluation of Series

Let $0<\delta<\pi$ and $f:\mathbb{R} \rightarrow \mathbb{R}$ be a $2\pi$ periodical function defined as:

$$f(x)= \left\{\begin{matrix}
1 &, &\left | x \right |\leq \delta \\
 0&, &\delta< \left | x \right |\leq \pi
\end{matrix}\right.$$

Expand $f$ in a Fourier series and in continuity show that:

$$\begin{matrix}
\displaystyle \sum_{n=1}^{\infty}\frac{\sin n \delta}{n}= \frac{\pi-\delta}{2} &, &\displaystyle \sum_{n=1}^{\infty}\frac{\sin^2 n\delta}{n^2 \delta}= \frac{\pi-\delta}{2}  &, &\displaystyle \sum_{n=1}^{\infty}\frac{1}{\left ( 2n-1 \right )^2}= \frac{\pi^2}{8}\end{matrix} $$

Solution



First of all we give the graph of $f$ which is a piecewise discontinuous function, but yet it can be expanded into a Fourier series.

So, we are computing the coefficients:

${\color{gray} \blacksquare} \;\; \displaystyle a_0= \frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\, {\rm d}t= \frac{1}{\pi}\left [ \cancelto{0}{\int_{-\pi}^{-\delta}0 \, {\rm d}t}+ \int_{-\delta}^{\delta}\, {\rm d}t+ \cancelto{0}{\int_{\delta}^{\pi}0 \, {\rm d}t}  \right ]=\frac{2\delta}{\pi}$

${\color{gray} \blacksquare} \;\; \displaystyle a_n= \frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\cos nt \, {\rm dt}= \frac{1}{\pi} \left [ \cancelto{0}{\int_{-\pi}^{-\delta}0 \, {\rm d}t} + \int_{-\delta}^{\delta}\cos nt \, {\rm d}t + \cancelto{0}{\int_{\delta}^{\pi} 0 \, {\rm d}t} \right ]=\\ \displaystyle \quad \quad \quad \qquad \qquad \qquad \qquad \;\;=\frac{2\sin \delta n}{\pi n}$

${\color{gray} \blacksquare} \;\; \displaystyle b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\sin nt \, {\rm d}t = 0$

Hence, the Fourier series of $f$ is:

$$f(x)= \frac{\delta}{\pi}+ \sum_{n=1}^{\infty}\frac{2\sin \delta n}{\pi n}\; \cos nx$$

Now we are in position, taking advantage of the fourier series of $f$ to evaluate the given three series.

1. $\displaystyle \sum_{n=1}^{\infty}\frac{\sin n\delta}{n}= \frac{\pi-\delta}{2}$

Proof:

 Setting $x=0$ in the identity involving the Fourier series we get that:

$$ 1= \frac{\delta}{\pi}+ \sum_{n=1}^{\infty}\frac{2\sin n\delta}{\pi n}\Rightarrow \sum_{n=1}^{\infty}\frac{\sin n \delta}{n} = \frac{\pi-\delta}{2}$$

2. $\displaystyle \sum_{n=1}^{\infty}\frac{\sin^2 n \delta}{n^2 \delta}= \frac{\pi-\delta}{2}$

 Proof:

We are making use of Parseval's identity , stating that:

$$\sum_{n=1}^{\infty}\left [ a_n^2 +b_n^2 \right ]+ \frac{a_0^2}{2}= \frac{1}{\pi}\int_{-\pi}^{\pi}f^2 (t)\, {\rm dt}$$

Hence:

$$ 4\sum_{n=1}^{\infty}\frac{\sin^2 n \delta}{\pi^2 n^2}+ \frac{2\delta^2}{\pi^2}= \frac{2\delta}{\pi}\Rightarrow \sum_{n=1}^{\infty}\frac{\sin^2 n \delta}{n^2 \delta}=  \frac{\pi-\delta}{2}$$

3. $\displaystyle \sum_{n=1}^{\infty}\frac{1}{\left ( 2n-1 \right )^2}= \frac{\pi^2}{8}$

Proof:

 In equation $2$ we substitute $\delta=\frac{\pi}{2}$ and we get the result since $\sin^2 \frac{\pi n}{2}$ is $1$ at odd numbers and $0$ at even numbers. Therefore we get the result.