Determine a série de Fourier de f(x) que supomos possuir o período 2 \pi.
f(x)=\left\{\begin{array}{lll}0 & \text { se } & -\pi<x<0 \\\\ 1 & \text { se } & 0<x<\frac{\pi}{2} \\\\ 0 & \text { se } & \frac{\pi}{2}<x<\pi\end{array}\right.

Passo 1

O enunciado pede a série de Fourier de período 2\pi da função
f(x)= \begin{cases} 0, & -\pi<x<0,\\\\ 1, & 0<x<\tfrac{\pi}{2},\\\\ 0, & \tfrac{\pi}{2}<x<\pi, \end{cases}
ou seja, vamos usar as fórmulas
a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\,dx,\quad a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)\,dx,\quad b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)\,dx
e montar
f(x)\sim \frac{a_0}{2}+\sum_{n=1}^{\infty}\bigl(a_n\cos(nx)+b_n\sin(nx)\bigr).

Passo 2

Cálculo de a_0:
a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\,dx =\frac{1}{\pi}\int_{0}^{\pi/2}1\,dx =\frac{1}{\pi}\cdot\frac{\pi}{2} =\frac12.

Passo 3

Cálculo de a_n para n\ge1:
a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)\,dx =\frac{1}{\pi}\int_{0}^{\pi/2}\cos(nx)\,dx =\frac{1}{\pi}\,\biggl[\frac{\sin(nx)}{n}\biggr]_{0}^{\pi/2} =\frac{\sin\bigl(n\pi/2\bigr)}{\pi\,n}.

Passo 4

Cálculo de b_n para n\ge1:
b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)\,dx =\frac{1}{\pi}\int_{0}^{\pi/2}\sin(nx)\,dx =\frac{1}{\pi}\,\biggl[-\frac{\cos(nx)}{n}\biggr]_{0}^{\pi/2} =\frac{1-\cos\bigl(n\pi/2\bigr)}{\pi\,n}.

Passo 5

Montando a série, com a_0/2=1/4:
f(x)\sim \frac14 +\sum_{n=1}^{\infty}\biggl( \frac{\sin\bigl(n\pi/2\bigr)}{\pi\,n}\cos(nx) +\frac{1-\cos\bigl(n\pi/2\bigr)}{\pi\,n}\sin(nx) \biggr).

Resposta

f(x)=\frac14 +\sum_{n=1}^{\infty}\biggl( \frac{\sin\bigl(n\pi/2\bigr)}{\pi\,n}\cos(nx) +\frac{1-\cos\bigl(n\pi/2\bigr)}{\pi\,n}\sin(nx) \biggr).

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