Example 4
Find the Fourier coefficients of:
$$
\begin{align*}
f(x)=\frac{\pi}{\sin(\pi\alpha)}e^{i(\pi-\theta)\alpha}
\end{align*}
$$
defined on \(0 \leq \theta \leq 2\pi\)
Let \(\alpha\notin\mathbb{Z}\) and define
$$
\begin{align*}
f(\theta)
&=
\frac{\pi}{\sin(\pi\alpha)}
e^{i(\pi-\theta)\alpha},
\qquad 0\leq\theta\leq 2\pi
\end{align*}
$$
Recall
$$
\begin{align*}
\widehat{f}(n)
&=
\frac{1}{2\pi}
\int_0^{2\pi}
f(\theta)e^{-in\theta}\,d\theta
\end{align*}
$$
Substitute the formula for \(f(\theta)\)
$$
\begin{align*}
\widehat{f}(n)
&=
\frac{1}{2\pi}
\int_0^{2\pi}
\left[
\frac{\pi}{\sin(\pi\alpha)}
e^{i(\pi-\theta)\alpha}
\right]
e^{-in\theta}\,d\theta \\
&=
\frac{\pi}{2\pi}
\int_0^{2\pi}
\frac{
e^{i(\pi-\theta)\alpha}e^{-in\theta}
}{
\sin(\pi\alpha)
}\,d\theta \\
&=
\frac{1}{2}
\int_0^{2\pi}
\frac{
e^{i(\pi-\theta)\alpha}e^{-in\theta}
}{
\sin(\pi\alpha)
}\,d\theta
\end{align*}
$$
Simplify the exponent:
$$
\begin{align*}
e^{i(\pi-\theta)\alpha}e^{-in\theta}
&=
e^{i\pi\alpha-i\theta\alpha}e^{-in\theta} \\
&=
e^{i\pi\alpha-i\theta\alpha-in\theta} \\
&=
e^{i\pi\alpha-i(\alpha+n)\theta} \\
&=
e^{i\pi\alpha}e^{-i(\alpha+n)\theta}
\end{align*}
$$
Therefore,
$$
\begin{align*}
\widehat{f}(n)
&=
\frac{1}{2}
\int_0^{2\pi}
\frac{
e^{i\pi\alpha}e^{-i(\alpha+n)\theta}
}{
\sin(\pi\alpha)
}\,d\theta \\
&=
\frac{e^{i\pi\alpha}}{2\sin(\pi\alpha)}
\int_0^{2\pi}
e^{-i(\alpha+n)\theta}\,d\theta
\end{align*}
$$
Integrate:
$$
\begin{align*}
\int
e^{-i(\alpha+n)\theta}\,d\theta
&=
\frac{
e^{-i(\alpha+n)\theta}
}{
-i(\alpha+n)
}
\end{align*}
$$
Hence
$$
\begin{align*}
\widehat{f}(n)
&=
\frac{e^{i\pi\alpha}}{2\sin(\pi\alpha)}
\left[
\frac{
e^{-i(\alpha+n)\theta}}{-i(\alpha+n)}
\right]_0^{2\pi} \\
&=
\frac{e^{i\pi\alpha}}{2\sin(\pi\alpha)}
\left[
\frac{
e^{-i(\alpha+n)2\pi}
}{
-i(\alpha+n)
}
-
\frac{
e^{-i(\alpha+n)0}
}{
-i(\alpha+n)
}
\right] \\
&=
\frac{e^{i\pi\alpha}}{2\sin(\pi\alpha)}
\left[
\frac{
e^{-i(\alpha+n)2\pi}
}{
-i(\alpha+n)
}
-
\frac{1}{-i(\alpha+n)}
\right] \\
&=
\frac{e^{i\pi\alpha}}{2\sin(\pi\alpha)}
\left[
\frac{
e^{-i(\alpha+n)2\pi}-1
}{
-i(\alpha+n)
}
\right] \tag{1}
\end{align*}
$$
Simplify \(e^{-i(\alpha+n)2\pi}\):
$$
\begin{align*}
e^{-i(\alpha+n)2\pi}
&=
e^{-i2\pi\alpha-i2\pi n} \\
&=
e^{-i2\pi\alpha}e^{-i2\pi n}
\end{align*}
$$
Since \(n\in\mathbb{Z}\), then \(e^{-i2\pi n} = \cos(2\pi n)-i\sin(2\pi n) = 1-0i = 1\). Thus
$$
\begin{align*}
e^{-i(\alpha+n)2\pi} &= e^{-i2\pi\alpha}e^{-i2\pi n} = e^{-i2\pi\alpha}(1) = e^{-i2\pi\alpha}
\end{align*}
$$
Substitute this back into (1)
$$
\begin{align*}
\widehat{f}(n)
&=
\frac{e^{i\pi\alpha}}{2\sin(\pi\alpha)}
\left[
\frac{
e^{-i2\pi\alpha}-1
}{
-i(\alpha+n)
}
\right] \\
&=
\frac{
e^{i\pi\alpha}
\left(e^{-i2\pi\alpha}-1\right)
}{
2\sin(\pi\alpha)\left[-i(\alpha+n)\right]
}
\end{align*}
$$
Distribute \(e^{i\pi\alpha}\) in the numerator:
$$
\begin{align*}
e^{i\pi\alpha}
\left(e^{-i2\pi\alpha}-1\right)
&=
e^{i\pi\alpha}e^{-i2\pi\alpha}
-
e^{i\pi\alpha} \\
&=
e^{i\pi\alpha-i2\pi\alpha}
-
e^{i\pi\alpha} \\
&=
e^{-i\pi\alpha}
-
e^{i\pi\alpha}
\end{align*}
$$
Therefore,
$$
\begin{align*}
\widehat{f}(n)
&=
\frac{
e^{-i\pi\alpha}-e^{i\pi\alpha}}{2\sin(\pi\alpha)\left[-i(\alpha+n)\right]} \tag{2}
\end{align*}
$$
Recall Euler’s formulas:
$$
\begin{align*}
e^{ix}
&=
\cos(x)+i\sin(x), \\
e^{-ix}
&=
\cos(x)-i\sin(x)
\end{align*}
$$
Hence,
$$
\begin{align*}
\widehat{f}(n)
&=
\frac{
-2i\sin(\pi\alpha)
}{
2\sin(\pi\alpha)\left[-i(\alpha+n)\right]
} \\
&=
\frac{
-2i\sin(\pi\alpha)
}{
-2i\sin(\pi\alpha)(\alpha+n)
} \\
&=
\frac{1}{\alpha+n}
\end{align*}
$$
Hence the Fourier coefficients are
$$
\begin{align*}
\widehat{f}(n)
&=
\frac{1}{n+\alpha},
\qquad n\in\mathbb{Z}
\end{align*}
$$
and the Fourier series is
$$
\begin{align*}
f(\theta)
&\sim
\sum_{n\in\mathbb{Z}}
\widehat{f}(n)e^{in\theta} =
\sum_{n\in\mathbb{Z}}
\frac{e^{in\theta}}{n+\alpha}
\end{align*}
$$
References
- SFSU Fourier Analysis by Professor Chun-Kit Lai
- Fourier Analysis: An Introduction by Elias M.Stein and Rami Shakarchi