Encontre a série de Fourier das funções, nos intervalos dados:
(d) f(x)=\left\{\begin{array}{cc}0, & -2 \leq x<0 \\\ x, & 0 \leq x<1 \\\\ 1, & 1 \leq x \leq 2\end{array}\right..

Passo 1

O enunciado nos dá a função periódica de período T=4\) definida em \([-2,2]\) por \[f(x)= \begin{cases} 0, & -2 \le x<0,\\\ x, & 0 \le x<1,\\\\ 1, & 1 \le x\le 2. \end{cases}
Queremos expandi-la em série de Fourier na forma
f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\Bigl(a_n\cos\bigl(\tfrac{n\pi x}{2}\bigr)+b_n\sin\bigl(\tfrac{n\pi x}{2}\bigr)\Bigr).
Vamos calcular a_0\], \(a_n e b_n usando as fórmulas com L=2:
a_0=\frac{1}{2}\int_{-2}^{2}f(x)\,dx,\quad a_n=\frac{1}{2}\int_{-2}^{2}f(x)\cos\Bigl(\tfrac{n\pi x}{2}\Bigr)\,dx,\quad b_n=\frac{1}{2}\int_{-2}^{2}f(x)\sin\Bigl(\tfrac{n\pi x}{2}\Bigr)\,dx.

Passo 2

Cálculo de a_0. Como em [-2,0) a função vale zero, só integramos em [0,1] e [1,2]:
\int_{-2}^{2}f(x)\,dx =\int_{0}^{1}x\,dx+\int_{1}^{2}1\,dx =\frac{1}{2}+1=\frac{3}{2}.
Logo
a_0=\frac{1}{2}\cdot\frac{3}{2}=\frac{3}{4} \quad\Longrightarrow\quad \frac{a_0}{2}=\frac{3}{8}.

Passo 3

Cálculo de a_n. Defina \alpha=\tfrac{n\pi}{2}. Então
a_n=\frac{1}{2}\Bigl(\int_{0}^{1}x\cos(\alpha x)\,dx+\int_{1}^{2}1\cdot\cos(\alpha x)\,dx\Bigr).
Usando integrais por partes e a primitiva de \cos(\alpha x), obtém-se
\int_{0}^{1}x\cos(\alpha x)\,dx =\frac{\sin\alpha}{\alpha}+\frac{\cos\alpha-1}{\alpha^2},
\int_{1}^{2}\cos(\alpha x)\,dx =\frac{\sin(2\alpha)-\sin\alpha}{\alpha}.
Somando e simplificando, os termos em \sin\alpha/\alpha cancelam, restando
a_n =\frac{1}{2}\Bigl(\frac{\cos\alpha-1}{\alpha^2}+\frac{\sin(2\alpha)}{\alpha}\Bigr).

Passo 4

Cálculo de b_n. Com a mesma \alpha=\tfrac{n\pi}{2}:
b_n=\frac{1}{2}\Bigl(\int_{0}^{1}x\sin(\alpha x)\,dx+\int_{1}^{2}\sin(\alpha x)\,dx\Bigr).
Calcula-se
\int_{0}^{1}x\sin(\alpha x)\,dx =-\frac{\cos\alpha}{\alpha}+\frac{\sin\alpha}{\alpha^2},
\int_{1}^{2}\sin(\alpha x)\,dx =\frac{\cos\alpha-\cos(2\alpha)}{\alpha}.
Os termos em \cos\alpha/\alpha se cancelam, dando
b_n =\frac{1}{2}\Bigl(\frac{\sin\alpha}{\alpha^2}-\frac{\cos(2\alpha)}{\alpha}\Bigr).

Passo 5

Montagem da série de Fourier. Com \alpha=n\pi/2, temos finalmente
f(x) =\frac{3}{8} +\sum_{n=1}^{\infty}\Bigl[ a_n\cos\bigl(\tfrac{n\pi x}{2}\bigr) +b_n\sin\bigl(\tfrac{n\pi x}{2}\bigr) \Bigr],
onde
a_n =\frac{1}{2}\Bigl(\frac{\cos\bigl(\tfrac{n\pi}{2}\bigr)-1}{(\tfrac{n\pi}{2})^2} +\frac{\sin\bigl(n\pi\bigr)}{\tfrac{n\pi}{2}}\Bigr), \quad b_n =\frac{1}{2}\Bigl(\frac{\sin\bigl(\tfrac{n\pi}{2}\bigr)}{(\tfrac{n\pi}{2})^2} -\frac{\cos\bigl(n\pi\bigr)}{\tfrac{n\pi}{2}}\Bigr).
Note que \sin(n\pi)=0 e \cos(n\pi)=(-1)^n, o que pode simplificar ainda mais.

Resposta

f(x) =\frac{3}{8} +\sum_{n=1}^{\infty}\Bigl[ \frac{1}{2}\Bigl(\frac{\cos\!\tfrac{n\pi}{2}-1}{(\tfrac{n\pi}{2})^2}\Bigr) \cos\!\bigl(\tfrac{n\pi x}{2}\bigr) +\frac{1}{2}\Bigl(\frac{\sin\!\tfrac{n\pi}{2}}{(\tfrac{n\pi}{2})^2} -\frac{(-1)^n}{\tfrac{n\pi}{2}}\Bigr) \sin\!\bigl(\tfrac{n\pi x}{2}\bigr) \Bigr].

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