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

Passo 1

O enunciado define a função
f(x)= \begin{cases} 0,& -\tfrac\pi2<x<0,\\\ \cos x,&0\le x<\tfrac\pi2 \end{cases}
no intervalo -\tfrac\pi2<x<\tfrac\pi2. O conteúdo é série de Fourier de período 2L=\pi, logo L=\tfrac\pi2. Usaremos as fórmulas
a_0=\frac{1}{L}\int_{-L}^{L}f(x)\,dx,
a_n=\frac{1}{L}\int_{-L}^{L}f(x)\cos\!\bigl(\tfrac{n\pi x}{L}\bigr)\,dx,
b_n=\frac{1}{L}\int_{-L}^{L}f(x)\sin\!\bigl(\tfrac{n\pi x}{L}\bigr)\,dx.

Passo 2

Como f(x)=0 para x<0,
a_0=\frac{1}{L}\int_{-L}^{L}f(x)\,dx =\frac{1}{\frac\pi2}\int_{0}^{\frac\pi2}\cos x\,dx =\frac{2}{\pi}\bigl[\sin x\bigr]_{0}^{\frac\pi2} =\frac{2}{\pi}\cdot1 =\frac{2}{\pi}.
No desenvolvimento da série aparece a_0/2=1/\pi.

Passo 3

Para a_n note que \tfrac{n\pi x}{L}=2n x, então
a_n =\frac{1}{L}\int_{-L}^{L}f(x)\cos(2n x)\,dx =\frac{2}{\pi}\int_{0}^{\frac\pi2}\cos x\cos(2n x)\,dx.
Usando
\cos x\cos(2n x) =\tfrac12\bigl[\cos((2n-1)x)+\cos((2n+1)x)\bigr]
temos
\int_{0}^{\frac\pi2}\cos x\cos(2n x)\,dx =\frac12\Bigl[\frac{\sin((2n-1)x)}{2n-1}+\frac{\sin((2n+1)x)}{2n+1}\Bigr]_{0}^{\frac\pi2} =\frac{(-1)^{n+1}}{4n^2-1}.
Logo
a_n =\frac{2}{\pi}\,\frac{(-1)^{n+1}}{4n^2-1} =\frac{2(-1)^{n+1}}{\pi(4n^2-1)}.

Passo 4

Para b_n escrevemos
b_n =\frac{1}{L}\int_{-L}^{L}f(x)\sin(2n x)\,dx =\frac{2}{\pi}\int_{0}^{\frac\pi2}\cos x\sin(2n x)\,dx.
Com
\cos x\sin(2n x) =\tfrac12\bigl[\sin((2n+1)x)+\sin((2n-1)x)\bigr]
segue
\int_{0}^{\frac\pi2}\cos x\sin(2n x)\,dx =\frac12\Bigl[-\frac{\cos((2n+1)x)}{2n+1}-\frac{\cos((2n-1)x)}{2n-1}\Bigr]_{0}^{\frac\pi2} =\frac{1}{2}\Bigl(\frac{1}{2n+1}+\frac{1}{2n-1}\Bigr).
Assim
b_n =\frac{2}{\pi}\cdot\frac{1}{2}\Bigl(\frac{1}{2n+1}+\frac{1}{2n-1}\Bigr) =\frac{1}{\pi}\Bigl(\frac{1}{2n+1}+\frac{1}{2n-1}\Bigr) =\frac{4n}{\pi(4n^2-1)}.

Passo 5

Reunindo, a série de Fourier de f em -\tfrac\pi2<x<\tfrac\pi2 é
f(x) =\frac{a_0}{2} +\sum_{n=1}^{\infty} \Bigl[a_n\cos(2n x)+b_n\sin(2n x)\Bigr]
com
\frac{a_0}{2}=\frac{1}{\pi},\quad a_n=\frac{2(-1)^{n+1}}{\pi(4n^2-1)},\quad b_n=\frac{4n}{\pi(4n^2-1)}.

Resposta

f(x) =\frac{1}{\pi} +\sum_{n=1}^{\infty} \Biggl[ \frac{2(-1)^{n+1}}{\pi(4n^2-1)}\cos(2n x) +\frac{4n}{\pi(4n^2-1)}\sin(2n x) \Biggr].

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