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