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In this Chapter we consider Fourier transform which is the most useful of all integral transforms.
In Section 4.5 we wrote Fourier series in the complex form \begin{equation} f(x)= \sum_{n=-\infty}^\infty c_n e^{\frac{i\pi nx}{l}} \label{eq-5.1.1} \end{equation} with \begin{equation} c_n= \frac{1}{2l}\int_{-l}^l f(x)e^{-\frac{i\pi n x}{l}}\,dx \qquad n=\ldots,-2, -1, 0,1,2,\ldots \label{eq-5.1.2} \end{equation} and \begin{equation} 2l\sum_{n=-\infty}^\infty |c_n|^2=\int_{-l}^l|f(x)|^2\,dx. \label{eq-5.1.3} \end{equation}
From this form we formally without any justification deduct Fourier integral.
First we introduce \begin{equation} k _n := \frac{\pi n}{l}\qquad \text{and}\qquad \Delta k _n = k _{n}- k _{n-1}= \frac{\pi}{l} \label{eq-5.1.4} \end{equation} and rewrite (\ref{eq-5.1.1}) as \begin{equation} f(x)= \sum_{n=-\infty}^\infty C( k _n) e^{i k _n x}\Delta k _n \label{eq-5.1.5} \end{equation} with \begin{equation} C( k )= \frac{1}{2\pi}\int_{-l}^l f(x)e^{-i k x}\,dx \label{eq-5.1.6} \end{equation} where we used $C( k _n) := c_n /(\Delta k _n)$; (\ref{eq-5.1.3}) should be rewritten as \begin{equation} \int_{-l}^l|f(x)|^2\,dx= 2\pi\sum_{n=-\infty}^\infty |C( k _n)|^2\Delta k _n. \label{eq-5.1.7} \end{equation} Now we formally set $l\to +\infty$; then integrals from $-l$ to $l$ in the right-hand expression of (\ref{eq-5.1.6}) and the left-hand expression of (\ref{eq-5.1.7}) become integrals from $-\infty$ to $+\infty$.
Meanwhile, $\Delta k _n \to +0$ and Riemannian sums in the right-hand expressions of (\ref{eq-5.1.5}) and (\ref{eq-5.1.7}) become integrals: \begin{equation} f(x)= \int_{-\infty}^\infty C( k ) e^{i k x}\,d k \label{eq-5.1.8} \end{equation} with \begin{equation} C( k )= \frac{1}{2\pi}\int_{-\infty}^\infty f(x)e^{-i k x}\,dx; \label{eq-5.1.9} \end{equation} (\ref{eq-5.1.3}) becomes \begin{equation} \int_{-\infty}^\infty |f(x)|^2\,dx= 2\pi\int_{-\infty}^\infty |C( k )|^2\,d k . \label{eq-5.1.10} \end{equation}
Definition 1. Formula (\ref{eq-5.1.9}) gives us a Fourier transform of $f(x)$, it usually is denoted by "hat": \begin{equation} \hat{f}( k )= \frac{1}{2\pi}\int_{-\infty}^\infty f(x)e^{-i k x}\,dx; \tag{FT}\label{FT} \end{equation} sometimes it is denoted by "tilde" ($\tilde{f}$), and seldom just by a corresponding capital letter $F( k )$.
Definition 2. Expression (\ref{eq-5.1.8}) is a Fourier integral aka inverse Fourier transform: \begin{equation} f(x)= \int_{-\infty}^\infty \hat{f}( k ) e^{i k x}\,d k \tag{FI}\label{FI} \end{equation} aka \begin{equation} \check{F}(x)= \int_{-\infty}^\infty F( k ) e^{i k x}\,d k \label{IFT}\tag{IFT} \end{equation}
Remark 1. Formula (\ref{eq-5.1.10}) is known as Plancherel theorem \begin{equation} \int_{-\infty}^\infty |f(x)|^2\,dx=2\pi\int_{-\infty}^\infty |\hat{f}( k )|^2\,d k . \tag{PT}\label{PT} \end{equation}
Remark 2. a. Sometimes expoments of $\pm i k x$ is replaced by $\pm 2\pi i k x$ and factor $1/(2\pi)$ dropped. b. Sometimes factor $\frac{1}{\sqrt{2\pi}}$ is placed in both Fourier transform and Fourier integral: \begin{align} &\hat{f}( k )= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{-i k x}\,dx; \tag{FT*}\\ &f(x)= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \hat{f}( k ) e^{i k x}\,d k \tag{FI*} \end{align} Then FT and IFT differ only by $i$ replaced by $-i$ and Plancherel theorem becomes \begin{equation} \int_{-\infty}^\infty |f(x)|^2\,dx= \int_{-\infty}^\infty |\hat{f}( k )|^2\,d k . \tag{PT*} \end{equation} In this case Fourier transform and inverse Fourier transform differ only by $-i$ instead of $i$ (very symmetric form) and both are unitary operators.
Remark 3. We can consider corresponding operator $LX=-X''$ in the space $L^2(\mathbb{R})$ of the square integrable functions on $\mathbb{R}$ but $e^{i k x}$ are no more eigenfunctions since they do not belong to this space. In advanced Real Analysis such functions often are referred as generalized eigenfunctions.
Remark 4.
Applying the same arguments as in Section 4.5 we can rewrite formulae (\ref{eq-5.1.8})--(\ref{eq-5.1.10}) as \begin{equation} f(x)= \int_0^\infty \bigl( A( k ) \cos( k x) +B( k ) \sin ( k x)\bigr) \,d k \label{eq-5.1.11} \end{equation} with \begin{align} & A( k )= \frac{1}{\pi}\int_{-\infty}^\infty f(x)\cos ( k x) \,dx, \label{eq-5.1.12}\\ & B( k )= \frac{1}{\pi}\int_{-\infty}^\infty f(x)\sin ( k x) \,dx, \label{eq-5.1.13} \end{align} and \begin{equation} \int_{-\infty}^\infty |f(x)|^2\,dx= \pi\int_0^\infty \bigl( |A( k )|^2+|B( k )|^2\bigr)\,d k . \label{eq-5.1.14} \end{equation}
$A( k )$ and $B( k )$ are $\cos$- and $\sin$- Fourier transforms and
Therefore
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