Example 1
Find the Fourier coefficients of:
$$
\begin{align*}
f(x) &=
\begin{cases}
1, & x \in [0,\pi],\\
-1, & x \in [-\pi,0].
\end{cases}
\end{align*}
$$
Recall that the Fourier coefficient is
$$
\begin{align*}
\widehat{f}(n) &= \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)e^{-inx}\ dx
\end{align*}
$$
We split the integral and substitute \(f(x)\)
$$
\begin{align*}
\widehat{f}(n) &= \frac{1}{2\pi}\int_{-\pi}^{0} f(x)e^{-inx}\ dx
+ \frac{1}{2\pi}\int_{0}^{\pi} f(x)e^{-inx}\ dx\\
&= \frac{1}{2\pi} \left( -\int_{-\pi}^{0}e^{-inx}\,dx +\int_{0}^{\pi}e^{-inx}\,dx \right)\\
&= \frac{1}{2\pi} \left( -\left[\frac{e^{-inx}}{-in}\right]_{-\pi}^{0} +\left[\frac{e^{-inx}}{-in}\right]_{0}^{\pi} \right)\\
&= \frac{1}{2\pi} \left( \frac{1}{in}\left(1-e^{in\pi}\right) +\frac{1}{-in}\left(e^{-in\pi}-1\right) \right). \tag{1}
\end{align*}
$$
Observe that
$$
\begin{align*}
e^{in\pi} &= \cos(n\pi) + i\sin(n\pi) = (-1)^n \\
e^{-in\pi} &= \cos(-n\pi) + i\sin(-n\pi)\\
&= \cos(n\pi) - i\sin(n\pi) = (-1)^n
\end{align*}
$$
Substituting back in (1), we obtain
$$
\begin{align*}
\widehat{f}(n) &= \frac{1}{2\pi} \left( \frac{1}{in}\left(1-(-1)^n\right) +\frac{1}{-in}\left((-1)^n-1\right)
\right) \\
&= \frac{1}{2\pi} \left( \frac{1}{in}\left(1-(-1)^n\right) +\frac{1}{in}\left(1-(-1)^n\right) \right)\\
&= \frac{1}{2\pi} \left( \frac{2}{in}\left(1-(-1)^n\right) \right).
\end{align*}
$$
Therefore,
$$
\begin{align*}
\widehat{f}(n) &= \begin{cases}
0, & n \text{ is even},\\
\dfrac{2}{\pi in}, & n \text{ is odd}.
\end{cases}
\end{align*}
$$
Thus,
$$
\begin{align*}
f(x) &\sim \sum_{\substack{n\in\mathbb{Z}\\ n\text{ odd}}} \frac{2}{\pi in}e^{inx}. \tag{2}
\end{align*}
$$
If we pair the positive and negative odd terms, for (n>0), then
$$
\begin{align*}
\frac{2}{\pi in}e^{inx} +\frac{2}{-\pi in}e^{-inx} &= \frac{2}{\pi in} \left(e^{inx}-e^{-inx}\right)
\end{align*}
$$
Recall now that
$$
\begin{align*}
e^{inx}-e^{-inx} = \cos(nx) + i\sin(nx) - (\cos(nx) - isin(nx))
&= 2i\sin(nx)
\end{align*}
$$
Hence
$$
\begin{align*}
\frac{2}{\pi in}e^{inx} +\frac{2}{-\pi in}e^{-inx} &= \frac{2}{\pi in}\left(2i\sin(nx)\right) = \frac{4}{\pi n}\sin(nx).
\end{align*}
$$
We can now re-write (2) as follows
$$
\begin{align*}
f(x)
&\sim
\sum_{\substack{n\ge 1\\ n\text{ odd}}}
\frac{4}{\pi n}\sin(nx)\\
&=
\frac{4}{\pi}
\left(
\sin x+\frac{\sin(3x)}{3}
+\frac{\sin(5x)}{5}+\cdots
\right).
\end{align*}
$$
References
- SFSU Fourier Analysis by Professor Chun-Kit Lai
- Fourier Analysis: An Introduction by Elias M.Stein and Rami Shakarchi