15.11 Obtenha a série de Fourier complexa de
f(x)=x+\mathrm{i}, \quad-\pi \leq x \leq+\pi, \quad \mathrm{i}=\sqrt{-1} .

Passo 1

O enunciado pede a série de Fourier complexa de
f(x)=x+\mathrm{i},\quad -\pi\le x\le\pi,
isto é, queremos escrever
f(x)=\sum_{n=-\infty}^{\infty}C_n\,e^{\mathrm{i}n x},
onde
C_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)\,e^{-\mathrm{i}n x}\,dx.

Passo 2

Começamos por n=0.
C_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}(x+\mathrm{i})\,dx =\frac{1}{2\pi}\Bigl[\underbrace{\int_{-\pi}^{\pi}x\,dx}_{0} +\mathrm{i}\int_{-\pi}^{\pi}1\,dx\Bigr] =\frac{1}{2\pi}(0+2\pi\,\mathrm{i}) =\mathrm{i}.

Passo 3

Agora n\neq0. Escrevemos
C_n=\frac{1}{2\pi}\Bigl(\int_{-\pi}^{\pi}x\,e^{-\mathrm{i}n x}\,dx +\mathrm{i}\int_{-\pi}^{\pi}e^{-\mathrm{i}n x}\,dx\Bigr).
A segunda integral é zero porque
\int_{-\pi}^{\pi}e^{-\mathrm{i}n x}\,dx =\Bigl[\frac{e^{-\mathrm{i}n x}}{-\mathrm{i}n}\Bigr]_{-\pi}^{\pi} =\frac{1}{-\mathrm{i}n}\bigl(e^{-\mathrm{i}n\pi}-e^{\mathrm{i}n\pi}\bigr) =\frac{(-1)^n-(-1)^n}{-\mathrm{i}n}=0.
Para a primeira usamos integração por partes com u=x, dv=e^{-\mathrm{i}n x}dx:
\int_{-\pi}^{\pi}x\,e^{-\mathrm{i}n x}\,dx =\Bigl[\frac{-x\,e^{-\mathrm{i}n x}}{\mathrm{i}n}\Bigr]_{-\pi}^{\pi} +\frac{1}{\mathrm{i}n}\int_{-\pi}^{\pi}e^{-\mathrm{i}n x}\,dx =-\frac{2\pi(-1)^n}{\mathrm{i}n} =\frac{2\pi\,\mathrm{i}(-1)^n}{n}.
Logo, para n\neq0
C_n=\frac{1}{2\pi}\cdot\frac{2\pi\,\mathrm{i}(-1)^n}{n} =\frac{\mathrm{i}(-1)^n}{n}.

Passo 4

Reunindo tudo, a expansão complexa é
f(x)=\sum_{n=-\infty}^{\infty}C_n\,e^{\mathrm{i}n x} =\mathrm{i}+\sum_{\substack{n=-\infty\n\neq0}}^{\infty}\frac{\mathrm{i}(-1)^n}{n}\,e^{\mathrm{i}n x}.

Resposta

f(x)=x+\mathrm{i} =\mathrm{i}+\sum_{\substack{n=-\infty\n\neq0}}^{\infty}\frac{\mathrm{i}(-1)^n}{n}\,e^{\mathrm{i}n x}.

Exercícios relacionados