$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$ #[Chapter 5. Fourier transform](id:chapter-5) In this Chapter we consider Fourier transform which is the most useful of all integral transforms. ##[5.1. Fourier transform, Fourier integral](id:sect-5.1) ----------------------------------- > 1. [Heuristics](#sect-5.1.1) > 2. [Definitions and Remarks](#sect-5.1.2) > 3. [$\cos $- and $\sin$-Fourier transform and > integral](#sect-5.1.3) ###[Heuristics](id:sect-5.1.1) In [Section 4.5](../Chapter4/S4.5.html) 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} ###[Definitions and Remarks](id:sect-5.1.2) **[Definition 1.](id:definition-5.1.1)** 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.](id:definition-5.1.2)** 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.](id:remark-5.1.1)** 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.](id:remark-5.1.2)** 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.](id:remark-5.1.3)** 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.](id:remark-5.1.4)** a. For justification see [Subsection 5.1.A](./S5.1.A.html); b. Pointwise convergence is discussed in [Subsection 5.1.B](./S5.1.B.html); c. Multidimensional Fourier transform and Fourier integral are discussed in [Subsection 5.2.A](./S5.2.A.html). ###[$\cos $- and $\sin$-Fourier transform and integral](id:sect-5.1.3) Applying the same arguments as in [Section 4.5](../Chapter4/S4.5.html) 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 1. $f(x)$ is even function iff $B( k )=0$; 2. $f(x)$ is odd function iff $A( k )=0$. Therefore 1. Each function on $[0,\infty)$ ] could be decomposed into $\cos$-Fourier integral \begin{equation} f(x)= \int\_0^\infty A( k ) \cos ( k x) \,d k \label{eq-5.1.15} \end{equation} with \begin{equation} A( k )=\frac{2}{\pi}\int\_0^\infty f(x)\cos ( k x) \,dx. \label{eq-5.1.16} \end{equation} 2. Each function on $[0,\infty)$ ] could be decomposed into $\sin$-Fourier integral \begin{equation} f(x)= \int\_0^\infty B( k ) \sin ( k x) \,d k \label{eq-5.1.17} \end{equation} with \begin{equation} B( k )= \frac{2}{\pi}\int\_0^\infty f(x)\sin ( k x) \,dx. \label{eq-5.1.18} \end{equation} --------------- [$\Leftarrow$](../Chapter4/S4.5.html)  [$\Uparrow$](../contents.html)  [$\Downarrow$](./S5.1.A.html)  [$\Downarrow$](./S5.1.B.html)  [$\Rightarrow$](./S5.2.html)