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\noindent
Then the Fourier series ($*$) becomes
\[ f(x) = \sum_{n = -\infty}^\infty \frac{\Delta
k}{2\pi} e^{i k_n x} \int_{-L}^L f(x') e^{-i k_n
x'} \dd x' \]
But $\sum_{n = -\infty}^\infty \Delta kg(k_n) \to
\int_{-\infty}^\infty g(k) \dd k$ with $g(k_n) =
\frac{e^{i k_n x}}{2\pi} \int_{-L}^L f(x')
e^{-ikx'} \dd x'$. So take limit as $L \to \infty$
and we have
\[ f(x) = \frac{1}{2\pi} \int_{-\infty}^\infty
\dd k e^{ikx} \left[ \int_{-\infty}^\infty f(x')
e^{-ikx'} \dd x' \right] =
\mathcal{F}^{-1}(\mathcal{F}(f))(x) \]
(i.e. equation (8.3))
\[ \text{resolution $\mathcal{F}$ (decomposition)}
\to \text{synthesis (reconstruction)} \]
Note when $f(x)$ is discontinuous at $x$ (like
Fourier series) the Fourier Transform gives
\[ \mathcal{F}^{-1}(\mathcal{F}(f))(x) = \half
(f(x_-) + f(x_+)) \tag{8.7} \]

\subsection{Fourier Transform Properties}

\[ \tilde{f}(t) = \int_{-\infty}^\infty f(x)
e^{-ikx} \dd x \]
\begin{enumerate}[(1)]
    \item Linearity:
        \[ h(x) = \lambda f(x) + \mu g(x) \iff
        \tilde{h}(k) = \lambda \tilde{f}(k) + \mu
        \tilde{g}(k) \tag{8.8} \]
    \item Translation:
        \[ h(x) = f(x - \lambda) \iff \tilde{h}(k)
        = e^{-i\lambda k} \tilde{f}(k)
        \tag{8.9} \]
        \[ \tilde{h}(k) = \int f(x - \lambda)
        e^{-ikx} \dd x = \int f(y) e^{-ik(y +
        \lambda)} \dd y = e^{-i\lambda k}
        \tilde{f}(k) \]
    \item Frequency:
        \[ h(x) = e^{i\lambda x} f(x) \iff
        \tilde{h}(k) = \tilde{f}(k - \lambda)
        \tag{8.10} \]
    \item Scaling:
        \begin{flashcard}[scaling-property-of-FT]
        \prompt{What is the scaling property of the
        Fourier Transform?}
        \[ h(x) = \cloze{f(\lambda x)} \iff
        \tilde{h}(k) =
        \cloze{\frac{1}{|\lambda|} \tilde{f} \left(
        \frac{k}{\lambda} \right)} \tag{8.11} \]
        \end{flashcard}
        ($|\lambda|$ because $x \to -x$ changes
        limits)
    \item Multiplication by $x$:
        \begin{flashcard}[FT-multiplied-by-x]
        \prompt{What is the Fourier Transform of $xf(x)$?}
        \[ h(x) = xf(x) \iff \tilde{h}(k) =
        \cloze{i\tilde{f}'(k)} \tag{8.12} \]
        \end{flashcard}
        because
        \begin{align*}
            \int_{-\infty}^\infty xf(x) e^{-ikx}
            \dd x
            &= -\frac{1}{i} \dfrac{}{k}
            \left(\int_{-\infty}^\infty f(x) e^{-ikx}
            \dd x \right)
        \end{align*}
    \item Derivative:
        \begin{flashcard}[FT-of-derivative]
        \prompt{What is the Fourier Transform of
        $f'(x)$?}
        \[ \boxed{h(x) = f'(x) \iff \tilde{h}(k) =
        \cloze{ik \tilde{f}(k)}} \tag{8.13} \]
        \end{flashcard}
        because
        \begin{align*}
            \tilde{h}(k)
            &= \int_{-\infty}^\infty f'(x)
            e^{-ikx} \dd x \\
            &= [f(x) e^{-ikx}]_{-\infty}^\infty +
            ik\int_{-\infty}^\infty f(x) e^{-ikx}
            \dd x \\
            &= ik \tilde{f}(k)
        \end{align*}
    \item General duality: Consider (8.2) with $x
        \to -x$
        \[ f(-x) = \frac{1}{2\pi}
        \int_{-\infty}^\infty \tilde{f}(k)
        e^{-ikx} \dd k \]
        so $k \leftrightarrow x$
        \[ \implies f(-k) = \frac{1}{2\pi}
        \int_{-\infty}^\infty \tilde{f}(x) e^{ikx}
        \dd x \]
        Thus
        \begin{flashcard}[FT-duality]
        \[ \boxed{g(x) = \tilde{f}(x) \iff
        \tilde{g}(x) = \cloze{2\pi f(-k)}} \tag{8.14} \]
        \end{flashcard}
        We have $f(-x) = \frac{1}{2\pi}
        \mathcal{F}(\tilde{f})(x) = \frac{1}{2\pi}
        \mathcal{F}^2(f)(x)$, so repeating,
        $\mathcal{F}^4(f)(x) = 4\pi^2 f(x)$.
\end{enumerate}
Exercise: Verify 1-7. \myskip
``Top hat'' example:
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/b23ab0b6640811ed.png}
\end{center}
Find Fourier Transform for
\[ f(x) = \begin{cases}
    1 & |x| \le a \\
    0 & |x| > a
\end{cases} \]
($a > 0$)
\begin{align*}
    \tilde{f}(k)
    &= \int_{-\infty}^\infty f(x) e^{-ikx} \dd x
    \\
    &= \int_{-a}^a \cos kx \dd x \\
    &= 2 \frac{\sin ka}{a} \tag{8.15}
\end{align*}
\begin{hiddenflashcard}[FT-of-top-hat]
\prompt{What is the Fourier Transform of a top
hat function?}
\[ f(x) = \cloze{
\begin{cases}
    1 & |x| \le a \\
    0 & |x| > a
\end{cases}
}
\]
\[ \tilde{f}(k) = \cloze{
2 \frac{\sin ka}{a}
} \]
\end{hiddenflashcard}
Fourier inversion, then (8.3) implies
\[ \frac{1}{\pi} \int_{-\infty}^\infty e^{ikx}
\frac{\sin ka}{k} \dd k = \begin{cases}
    1 & |x| \le a \\
    0 & |x| > a
\end{cases} \]
Now set $x = 0$, then take $k \to x$ to obtain
Dirchlet discontinuous formula:
\begin{flashcard}[Dirchlet-discontinuous-formula]
\prompt{What is the Dirchlet discontinuous
formula?}
\[ \cloze{\int_0^\infty \frac{\sin ax}{x}
\dd x} \fcscrap{= \begin{cases}
    \frac{\pi}{2} & a > 0 \\
    0 & a = 0 \\
    -\frac{\pi}{2} & a < 0
\end{cases}}
= \cloze{\frac{\pi}{2} \sgn(a)} \tag{8.16} \]
\end{flashcard}
Here, we allow $a < 0$, so $\sin(-ax) = -\sin ax$.
(See RJ notes for direct inverse Fourier transform
of (8.15))

\subsection{Convolution and Parseval's Theorem}

We want to multiply Fourier Transforms in
frequency domain $\tilde{h}(k) = \tilde{f}(k)
\tilde{g}(k)$ so consider the inverse:
\begin{align*}
    h(x)
    &= \frac{1}{2\pi} \int_{-\infty}^\infty
    \tilde{f}(k) \tilde{g}(k) e^{ikx} \dd k \\
    &= \frac{1}{2\pi} \int_{-\infty}^\infty \left(
    \int_{-\infty}^\infty f(y) e^{-iky} \dd
    y\right) \tilde{g}(k) e^{ikx} \dd k \\
    &= \int_{-\infty}^\infty f(y) \left(
    \frac{1}{2\pi} \int_{-\infty}^\infty
    \tilde{g}(k) e^{ik(x - y)} \dd k \right) \dd y
    &&\text{see (8.9)} \\
    &= \int_{-\infty}^\infty f(y) g(x - y) \dd y
    \\
    &\equiv f * g(x) \tag{8.17}
\end{align*}
(convlution definition). By duality (8.14) we also
have
\begin{flashcard}[FT-of-product]
\prompt{Fourier transform of product?}
\[ h(x) = f(x)g(x) \iff \tilde{h}(k) =
\cloze{\frac{1}{2\pi} \int_{-\infty}^\infty
\tilde{f}(p) \tilde{g}(k - p) \dd p} \tag{8.18} \]
\prompt{\cloze{(the LHS is $\frac{1}{2\pi}
\tilde{f} * \tilde{g} (k)$).}}
\end{flashcard}

\subsubsection*{Parseval's Theorem}

Consider $h(x) = g^*(-x)$, then
\begin{align*}
    \tilde{h}(k)
    &= \int_{-\infty}^\infty g^*(-x) e^{-ikx} \dd
    x \\
    &= \left[ \int_{-\infty}^\infty g(-x) e^{ikx}
    \dd x \right]^* \\
    &= \left[ \int_{-\infty}^\infty g(y) e^{-iky}
    \dd y \right]^* \\
    &= \tilde{g}^*(k)
\end{align*}
Substitute into (8.17) $g(x) \to g^*(-x)$,
\[ \int_{-\infty}^\infty f(y) g^*(y - x) \dd y =
\frac{1}{2\pi} \int_{-\infty}^\infty \tilde{f}(k)
\tilde{g}^*(k) e^{ikx} \dd k \]
Take $x = 0$, then dummy variable $y \to x$ on
LHS:
\[ \int_{-\infty}^\infty f(x) g^*(x) \dd x =
\frac{1}{2\pi} \int_{-\infty}^\infty \tilde{f}(k)
\tilde{g}^*(k) \dd k \tag{8.19} \]
Or equivalently,
\[ \langle g, f \rangle = \frac{1}{2\pi} \langle
\tilde{g}, \tilde{f} \rangle_k \]
see section (2.1). Now $g = f^*$:
\begin{flashcard}[FT-Parseval-theorem]
\prompt{Parseval's theorem for FT?}
\[ \boxed{\cloze{\int_{-\infty}^\infty |f(x)|^2
\dd x} = \cloze{
\frac{1}{2\pi} \int_{-\infty}^\infty
|\tilde{f}(k)|^2 \dd k}} \tag{8.20} \]
\end{flashcard}
which is \emph{Parseval's theorem}.
