Determine a série de Fourier de f no intervalo indicado.
f(x)=\left\{\begin{array}{lr}\pi^{2}, & -\pi<x<0 \\\ \pi^{2}-x^{2}, & 0 \leq x<\pi\end{array}\right.

Passo 1

O enunciado pede a série de Fourier de
f(x)= \begin{cases} \pi^{2}, & -\pi < x < 0,\\\ \pi^{2}-x^{2}, & 0 \le x < \pi \end{cases}
no intervalo (-\pi,\pi).
O que precisamos é usar a forma geral da série de Fourier em (-\pi,\pi):
f(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}\bigl(a_{n}\cos(nx)+b_{n}\sin(nx)\bigr),
onde
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.

Passo 2

Cálculo de a_{0}:
\int_{-\pi}^{\pi}f(x)\,dx =\int_{-\pi}^{0}\pi^{2}\,dx+\int_{0}^{\pi}(\pi^{2}-x^{2})\,dx =\pi^{2}\cdot \pi+\Bigl(\pi^{2}\pi-\frac{\pi^{3}}{3}\Bigr) =2\pi^{3}-\frac{\pi^{3}}{3}=\frac{5\pi^{3}}{3}.
Logo
a_{0}=\frac{1}{\pi}\,\frac{5\pi^{3}}{3}=\frac{5\pi^{2}}{3},
e
\frac{a_{0}}{2}=\frac{5\pi^{2}}{6}.

Passo 3

Cálculo de a_{n} para n\ge1:
a_{n} =\frac{1}{\pi}\Bigl[ \int_{-\pi}^{0}\pi^{2}\cos(nx)\,dx +\int_{0}^{\pi}(\pi^{2}-x^{2})\cos(nx)\,dx \Bigr].
Primeiro termo:
\int_{-\pi}^{0}\pi^{2}\cos(nx)\,dx =\pi^{2}\Bigl[\frac{\sin(nx)}{n}\Bigr]_{-\pi}^{0} =\pi^{2}\frac{\sin(0)-\sin(-n\pi)}{n} =0.
Fica
a_{n} =\frac{1}{\pi}\int_{0}^{\pi}(\pi^{2}-x^{2})\cos(nx)\,dx.
Separando,
\int_{0}^{\pi}\pi^{2}\cos(nx)\,dx=0, \quad \int_{0}^{\pi}x^{2}\cos(nx)\,dx =\frac{2\pi(-1)^{n}}{n^{2}}
(por integração por partes). Portanto
\int_{0}^{\pi}(\pi^{2}-x^{2})\cos(nx)\,dx =-\frac{2\pi(-1)^{n}}{n^{2}},
a_{n} =\frac{1}{\pi}\Bigl(-\frac{2\pi(-1)^{n}}{n^{2}}\Bigr) =-\frac{2(-1)^{n}}{n^{2}}.

Passo 4

Cálculo de b_{n} para n\ge1:
b_{n} =\frac{1}{\pi}\Bigl[ \int_{-\pi}^{0}\pi^{2}\sin(nx)\,dx +\int_{0}^{\pi}(\pi^{2}-x^{2})\sin(nx)\,dx \Bigr].
Mostra-se que as parcelas com \pi^{2}\sin(nx) se cancelam, restando
b_{n} =-\frac{1}{\pi}\int_{0}^{\pi}x^{2}\sin(nx)\,dx.
Por duas integrações por partes chega-se a
\int_{0}^{\pi}x^{2}\sin(nx)\,dx =-\frac{\pi^{2}(-1)^{n}}{n}+\frac{2\bigl((-1)^{n}-1\bigr)}{n^{3}},
logo
b_{n} =-\frac{1}{\pi}\Bigl(-\frac{\pi^{2}(-1)^{n}}{n} +\frac{2\bigl((-1)^{n}-1\bigr)}{n^{3}}\Bigr) =\frac{\pi(-1)^{n}}{n}-\frac{2\bigl((-1)^{n}-1\bigr)}{\pi\,n^{3}}.
Uma forma compacta é
b_{n} =(-1)^{n}\Bigl(\frac{\pi}{n}-\frac{2}{\pi\,n^{3}}\Bigr) +\frac{2}{\pi\,n^{3}}.

Passo 5

Juntando tudo, a série de Fourier de f em (-\pi,\pi) é
f(x) =\frac{5\pi^{2}}{6} +\sum_{n=1}^{\infty}\Bigl[ -\frac{2(-1)^{n}}{n^{2}}\cos(nx) +\Bigl((-1)^{n}\Bigl(\frac{\pi}{n}-\frac{2}{\pi\,n^{3}}\Bigr) +\frac{2}{\pi\,n^{3}}\Bigr)\sin(nx) \Bigr].

Resposta

f(x) =\frac{5\pi^{2}}{6} +\sum_{n=1}^{\infty}\biggl[ -\frac{2(-1)^{n}}{n^{2}}\cos(nx) +\Bigl( (-1)^{n}\Bigl(\frac{\pi}{n}-\frac{2}{\pi\,n^{3}}\Bigr) +\frac{2}{\pi\,n^{3}} \Bigr)\sin(nx) \biggr].

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