Complex version of Fourier series

Question

Let a be a positive real number and f $2\pi$ periodic function defined by:

$f(x)= \begin{cases}0 & \text{if }-\pi\lt x\lt 0 \\ 1 &\text{if }0\le x\le a \\ 0 &\text{if }a\lt x\le \pi\end{cases}$

Find the complex form of Fourier series of $f(x)$

Answer 1

The complex form of the Fourier series associated to $f$ is of the form:

$$\sum_{n \in \mathbb Z} c_n e^{inx}$$

Where:

$$c_n = \begin{cases} \frac{a_n - ib_n}2 &\text{if $n \ge 1$} \\ \frac{a_0}2 &\text{if $n = 0$} \\ \frac{a_{-n} + ib_{-n}}2 &\text{if $n \le -1$} \end{cases}$$

We just have to find the $a_n$'s and $b_n$'s.

As $f = 0$ on $(-\pi, 0)$ and $(a, \pi]$, our integrals are just over $[0,a]$.

$$a_0 = \frac1{\pi} \int_0^a (1) dx = \frac{a}{\pi}$$

For $n > 0$, we have:

$$a_n = \frac{1}{\pi} \int_0^a \cos(nx)dx = \frac{\sin(na)}{n \pi}$$

$$b_n = \frac{1}{\pi} \int_0^a \sin(nx) dx = \frac{1 - \cos(na)}{n \pi}$$

Therefore:

$$c_n = \begin{cases} i\frac{e^{-ina} - 1}{2\pi n} &\text{if $n \ge 1$} \\ \frac{a}{2\pi} &\text{if $n = 0$} \\ i\frac{e^{-ina} - 1}{2\pi n} &\text{if $n \le -1$} \end{cases}$$

Therefore:

$$ f(x) \sim \frac{a}{2 \pi} +\frac{i}{2\pi} \sum_{n \in \mathbb Z^*} \left( \frac{e^{-ina} - 1}{n} \right) e^{inx} $$