Fourier transform, Fourier integral

$\renewcommand{\Re}{\operatorname{Re}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$

Properties of Fourier transform

  1. Basic properties
  2. Convolution
  3. Examples

Basic properties

In the previous Lecture 17 we introduced Fourier transform and Inverse Fourier transform

\begin{align} & \hat{f}(\omega)= \frac{\kappa}{2\pi}\int_{-\infty}^\infty f(x)e^{-i\omega x}\,dx \tag{FT}\\ & \check{F}(x)= \frac{1}{\kappa} \int_{-\infty}^\infty F(\omega) e^{i\omega x}\,d \omega \tag{IFT} \end{align} with $\kappa=1$ (but here we will be a bit more flexible):

Theorem 1. $F= \hat{f} \iff f=\check{F}$.

(Already "proved")

Theorem 2.

  1. Fourier transform: $f\mapsto \hat{f}$ is a linear operator $L^2(\mathbb{R},\mathbb{C})\to L^2(\mathbb{R},\mathbb{C})$;
  2. Inverse Fourier transform: $F\mapsto \check{F}$ is an inverse operator (and also a linear operator) $L^2(\mathbb{R},\mathbb{C})\to L^2(\mathbb{R},\mathbb{C})$;
  3. If $\kappa=\frac{1}{\sqrt{2\pi}}$ these operators are unitary i.e. preserve norm and an inner product: \begin{gather} \|f\|=\Bigl(\int_{\mathbb{R}} |f(x)|^2\,dx\Bigr)^{\frac{1}{2}}, \label{equ-18.1}\\[5pt] (f,g)= \int_{\mathbb{R}} f(x)\bar{g}(x)\,dx. \label{equ-18.2} \end{gather}


  1. Here $L^2(\mathbb{R},\mathbb{C})$ is a space of square integrable complex-valued functions. Accurate definition requires a measure theory (studied in the course of Real Analysis). Alternatively one can introduce this space as a closure of the set of square integrable continuous functions but it also require a certain knowledge of Real Analysis.
  2. Properties (a) and (b) are obvious and (c) is due to Plancherel's theorem.

Theorem 3.

  1. $g(x)=f(x-a)\implies \hat{g}(\omega)=e^{-i\omega a}\hat{f}(\omega)$;

  2. $g(x)=f(x)e^{ibx}\implies \hat{g}(\omega)=\hat{f}(\omega-b)$;

  3. $g(x)=f'(x) \implies \hat{g}(\omega)=i\omega \hat{f}(\omega)$;

  4. $g(x)=xf (x) \implies \hat{g}(\omega)=i \hat{f}{}'(\omega)$;

  5. $g(x)=f(\lambda x)\implies \hat{g}(\omega)=|\lambda|^{-1} \hat{f}(\lambda^{-1}\omega)$;

Proof. Here for brevity we do not write that all integrals are over $\mathbb{R}$ and set $\kappa=2\pi$.

  1. $\hat{g}=\int e^{-i\omega x}g(x)\,dx = \int e^{-i \omega x}f(x-a)\,dx= \int e^{-i\omega(x+a)}f(x)\,dx= e^{-i\omega a}\hat{f}(\omega)$. We replaced $x$ by $(x+a)$ in the integral.

  2. $\hat{g}=\int e^{-i \omega x}g(x)\,dx = \int e^{-i\omega x}e^{ibx}f(x)\,dx= \int e^{-i (\omega-b)x}f(x)\,dx= \hat{f}(\omega-b)$.

  3. $\hat{g}=\int e^{-i \omega x}g(x)\,dx = \int e^{-i\omega x}f'(x)\,dx \overset{\text{by parts}}= \int \bigl(e^{-i \omega x}\bigr)'f(x)\,dx= i\omega \hat{f}(\omega)$.

  4. $\hat{g}=\int e^{-i \omega x}g(x)\,dx = \int e^{-i\omega x}xf(x)\,dx= \int i\partial_\omega \bigl(e^{-i \omega x}\bigr) \, f(x)\,dx= i\hat{f}{}'(\omega)$.

  5. $\hat{g}=\int e^{-i\omega x}g(x)\,dx = \int e^{-i \omega x}f(\lambda x )\,dx= \int e^{-i\omega |\lambda|^{-1}x}f(x)\,\lambda^{-1}dx= \lambda^{-1}\hat{f}(\lambda^{-1}\omega)$. Here we replaced $x$ by $\lambda^{-1}x$ in the integral and $|\lambda|^{-1}$ is an absolute value of Jacobian.

Corollary 1. $f$ is even (odd) iff $\hat{f}$ is even (odd).


Definition. Convolution of functions $f$ and $g$ is a function $f*g$: \begin{equation} (f*g)(x):=\int f(x-y)g(y)\,dy. \label{equ-18.3} \end{equation}

Theorem 4.

  1. $h=f*g\implies \hat{h}(\omega)=\frac{2\pi}{\kappa}\hat{f}(\omega)\hat{g}(\omega)$;

  2. $h(x)=f(x)g(x)$ $\implies \hat{h}=\kappa \hat{f}*\hat{g}$;


  1. \begin{equation*} \hat{h}(x)=\frac{\kappa}{2\pi} \int e^{-ix\omega}h(x)\,dx = \frac{\kappa}{2\pi} \iint e^{-ix\omega}f(x-y)g(y)\,dxdy; \end{equation*} replacing in the integral $x:=y+z$ we arrive to \begin{equation*} \frac{\kappa}{2\pi} \iint e^{-i(y+z)\omega}f(z)g(y)\,dzdy= \frac{\kappa}{2\pi} \int e^{-iz\omega}f(z)\,dz \times \int e^{-iy\omega}g(y)\,dz \end{equation*} which is equal to $\frac{2\pi}{\kappa}\hat{f}(\omega)\hat{g}(\omega)$.

  2. Similarly $\hat{f}*\hat{g}$ is a Fourier transform of $\frac{\kappa_1}{2\pi}fg$ where $\kappa_1=\frac{2\pi}{\kappa}$.


Example 1. Let $f(x)=e^{-\alpha x}$ as $x>0$ and $f(x)=0$ as $x<0$. Here $\Re \alpha>0$. \begin{equation*} \hat{f}(\omega)= \int_0^\infty e^{-(\alpha +i\omega )x}\,dx = -(\alpha +i\omega )^{-1}e^{-(\alpha +i\omega )x}\bigr|_{x=0}^{x=\infty}= (\alpha +i\omega )^{-1} \end{equation*} provided $\kappa=2\pi$.

In the general case $\hat{f}(\omega)= \frac{\kappa}{2\pi}(\alpha+i\omega )^{-1}$.

Example 2. Let $f(x)=e^{-\frac{\alpha}{2}x^2}$ with $\Re\alpha\ge 0$. Here even for $\Re \alpha=0$ F.t. exists as integrals are converging albeit not absolutely.

Note that $f'=\alpha x f$. Applying Fourier transform and Theorem 3 (c),(d) to the left, right we get $i\omega \hat{f}= -i\alpha \hat{f}'$; solving it we arrive to $\hat{f}=Ce^{-\frac{1}{2\alpha}\omega^2}$.

To find $C$ note that $C=\hat{f}(0)= \frac{\kappa}{2\pi}\int e^{-\frac{\alpha}{2}x^2}\,dx$ and for real $\alpha>0$ we make a change of variables $x=\alpha^{-\frac{1}{2}}z$ and arrive to $C=\frac{\kappa}{\sqrt{2\pi \alpha}}$ because $\int e^{-z^2/2}\,dz=\sqrt{2\pi}$. Therefore \begin{equation*} \hat{f}(\omega)= \frac{\kappa}{\sqrt{2\pi\alpha}}e^{-\frac{1}{2\alpha}\omega^2}. \end{equation*} Knowing complex variables one can justify it for complex $\alpha $ with $\Re\alpha\ge 0$; we take a correct branch of $\sqrt{\alpha}$ (condition $\Re\alpha\ge 0$ prevents going around origin). In particular, $(\pm i )^{\frac{1}{2}}=e^{\pm \frac{i\pi}{4}}$ and therefore for $\alpha=\pm i\beta $ with for $\beta>0$ we get $f=e^{\mp\frac{i}{2\beta }x^2}$ and \begin{equation*} \hat{f}(\omega)=\frac{\kappa}{2\sqrt{\pi\beta}} (1\mp i)e^{\pm\frac{i}{2\beta}\omega^2x}. \end{equation*}