Example 2
Find the Fourier coefficients of \(f(x) = x\) defined on \((-\pi,\pi)\)
Note again that if we were to integrate from \(0\) to \(2\pi\) instead, then we want to move \(x\in[\pi,2\pi]\) back into the original interval \([-\pi,\pi]\):
$$
\begin{align*}
x-2\pi\in[-\pi,0].
\end{align*}
$$
Since the function is \(2\pi\)-periodic,
$$
\begin{align*}
f(x)=f(x-2\pi).
\end{align*}
$$
The original formula is \(f(u)=u\), so
$$
\begin{align*}
f(x)=f(x-2\pi)=x-2\pi.
\end{align*}
$$
Therefore, on \([0,2\pi]\), the original sawtooth becomes
$$
\begin{align*}
\boxed{
f(x)=
\begin{cases}
x,&0\le x\le\pi,\\[4pt]
x-2\pi,&\pi < x \le 2\pi.
\end{cases}}
\end{align*}
$$
Solution
Note that the function \(f(x)=x\) is piecewise continuous on \([-\pi,\pi]\) so it’s going to look like a sawtooth
Recall now that the Fourier coefficient is
$$
\begin{align*}
\widehat{f}(n) &= \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)e^{-inx}\ dx
\end{align*}
$$
If we substitute the function, then
$$
\begin{align*}
\widehat{f}(n)
&=
\frac{1}{2\pi}
\int_{-\pi}^{\pi}
xe^{-inx}\,dx
\end{align*}
$$
When \(n\neq 0\), we need to use integration by parts as follows:
$$
\begin{align*}
u &= x,
&du &= dx, \\
dv &= e^{-inx}\,dx,
&v &= \frac{e^{-inx}}{-in}
\end{align*}
$$
and recall that \(\int udv = uv - \int vdu\). Hence
$$
\begin{align*}
\int_{-\pi}^{\pi}xe^{-inx}\,dx
&=
\left[
x\frac{e^{-inx}}{-in}
\right]_{-\pi}^{\pi}
-
\int_{-\pi}^{\pi}
\frac{e^{-inx}}{-in}\,dx
\end{align*}
$$
We evaluate the first term as follows:
$$
\begin{align*}
\left[
x\frac{e^{-inx}}{-in}
\right]_{-\pi}^{\pi}
&=
\frac{\pi e^{-in\pi}}{-in}
-
\frac{(-\pi)e^{-in(-\pi)}}{-in} \\
&=
\frac{\pi e^{-in\pi}}{-in}
-
\frac{-\pi e^{in\pi}}{-in} \\
&=
\frac{\pi e^{-in\pi}}{-in}
+
\frac{\pi e^{in\pi}}{-in} \\
&=
\frac{\pi e^{-in\pi}+\pi e^{in\pi}}{-in}
\end{align*}
$$
Recall that when \(n \in \mathbb{Z}\), then
$$
\begin{align*}
e^{in\pi} &= \left(e^{i\pi}\right)^n = (-1)^n \\
e^{-in\pi} &= \left(e^{-i\pi}\right)^n = (-1)^n
\end{align*}
$$
Equivalently, since \((-1)^n\) is either \(1\) or \(-1\),
Therefore,
$$
\begin{align*}
\left[
x\frac{e^{-inx}}{-in}
\right]_{-\pi}^{\pi}
&= \frac{\pi(-1)^n+\pi(-1)^n}{-in} = \frac{2\pi(-1)^n}{-in}
\tag{1}
\end{align*}
$$
While the integral is
$$
\begin{align*}
\int_{-\pi}^{\pi}
\frac{e^{-inx}}{-in}\,dx
&=
\frac{1}{-in}
\int_{-\pi}^{\pi}
e^{-inx}\,dx \\
&=
\frac{1}{-in}
\left[
\frac{e^{-inx}}{-in}
\right]_{-\pi}^{\pi} \\
&=
\frac{1}{-in}
\left[
\frac{e^{-in\pi}}{-in}
-
\frac{e^{-in(-\pi)}}{-in}
\right] \\
&=
\frac{1}{-in}
\left[
\frac{e^{-in\pi}}{-in}
-
\frac{e^{in\pi}}{-in}
\right] \\
&=
\frac{1}{-in}
\left[
\frac{e^{-in\pi}-e^{in\pi}}{-in}
\right] \\
&=
\frac{1}{-in}
\left[
\frac{(-1)^n-(-1)^n}{-in}
\right] \\
&=
\frac{1}{-in}
\left[
\frac{0}{-in}
\right] \\
&= 0
\tag{2}
\end{align*}
$$
Hence, by (1) and (2),
$$
\begin{align*}
\int_{-\pi}^{\pi}xe^{-inx}\,dx
&=
\left[
x\frac{e^{-inx}}{-in}
\right]_{-\pi}^{\pi}
-
\int_{-\pi}^{\pi}
\frac{e^{-inx}}{-in}\,dx =
\frac{2\pi(-1)^n}{-in}
\end{align*}
$$
Finally, we see that the Fourier coefficient when \(n\neq 0\) is
$$
\begin{align*}
\widehat{f}(n)
&=
\frac{1}{2\pi}
\int_{-\pi}^{\pi}
xe^{-inx}\,dx \\
&=
\frac{1}{2\pi}
\left[
\frac{2\pi(-1)^n}{-in}
\right] \\
&=
\frac{2\pi(-1)^n}{-2\pi in} \\
&=
\frac{(-1)^n}{-in},
\qquad n\neq 0
\end{align*}
$$
When \(n=0\)
$$
\begin{align*}
\widehat{f}(0)
&=
\frac{1}{2\pi}
\int_{-\pi}^{\pi}
f(x)e^{-i(0)x}\,dx \\
&=
\frac{1}{2\pi}
\int_{-\pi}^{\pi}
x e^0\,dx \\
&=
\frac{1}{2\pi}
\int_{-\pi}^{\pi}
x\,dx \\
&=
\frac{1}{2\pi}
\left[
\frac{x^2}{2}
\right]_{-\pi}^{\pi} \\
&=
\frac{1}{2\pi}
\left[
\frac{\pi^2}{2}
-
\frac{(-\pi)^2}{2}
\right] \\
&=
\frac{1}{2\pi}
\left[
\frac{\pi^2}{2}
-
\frac{\pi^2}{2}
\right] \\
&=
\frac{1}{2\pi}(0) \\
&= 0
\end{align*}
$$
This also follows because \(x\) is an odd function and the interval \([-\pi,\pi]\) is symmetric. Finally we can group the coefficients in the following:
$$
\begin{align*}
\widehat{f}(n)
&=
\begin{cases}
0, & n=0, \\[6pt]
\dfrac{(-1)^n}{-in}, & n\neq 0
\end{cases}
\end{align*}
$$
The Fourier series then becomes
$$
\begin{align*}
f(x)
&\sim
\sum_{n\in\mathbb{Z}}
\widehat{f}(n)e^{inx} \\
&=
\widehat{f}(0)
+
\sum_{\substack{n\in\mathbb{Z}\\n\neq 0}}
\widehat{f}(n)e^{inx} \\
&=
0+
\sum_{\substack{n\in\mathbb{Z}\\n\neq 0}}
\frac{(-1)^n}{-in}e^{inx} \\
&=
\sum_{\substack{n\in\mathbb{Z}\\n\neq 0}}
\frac{(-1)^n}{-in}e^{inx}
\end{align*}
$$
If we further pair the terms corresponding to \(n\) and \(-n\), then
$$
\begin{align*}
f(x)
&\sim
\sum_{n=1}^{\infty}
\left[
\widehat{f}(n)e^{inx}
+
\widehat{f}(-n)e^{-inx}
\right]
\end{align*}
$$
Note that
$$
\begin{align*}
\widehat{f}(-n) &= \frac{(-1)^{-n}}{-i(-n)} =
\frac{(-1)^n}{in}
\end{align*}
$$
Hence
$$
\begin{align*}
f(x)
&\sim
\sum_{n=1}^{\infty}
\left[
\frac{(-1)^n}{-in}e^{inx}
+
\frac{(-1)^n}{in}e^{-inx}
\right] \\
&=
\sum_{n=1}^{\infty}
\frac{(-1)^n}{in}
\left[
-e^{inx}+e^{-inx}
\right]
\end{align*}
$$
Using
$$
\begin{align*}
e^{inx}-e^{-inx}
&=
2i\sin(nx)
\end{align*}
$$
We get:
$$
\begin{align*}
f(x)
&\sim
\sum_{n=1}^{\infty}
\frac{(-1)^n}{in}
\left[
-2i\sin(nx)
\right] \\
&=
\sum_{n=1}^{\infty}
\frac{-2i(-1)^n}{in}
\sin(nx) \\
&=
\sum_{n=1}^{\infty}
\frac{-2(-1)^n}{n}
\sin(nx) \\
&=
2\sum_{n=1}^{\infty}
\frac{(-1)^{n+1}}{n}\sin(nx)
\end{align*}
$$
References
- SFSU Fourier Analysis by Professor Chun-Kit Lai
- Fourier Analysis: An Introduction by Elias M.Stein and Rami Shakarchi