5.2B. Fourier transform in the complex domain

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5.2B. Fourier transform in the complex domain

  1. Introduction
  2. Paley-Wiener theorem
  3. Laplace transform
  4. Using complex variables


In this Appendix the familiarity with elements of the Complex Variables (like MAT334 at University of Toronto) is assumed.

When we can take in the definition \begin{equation} \hat{u}(z)= \frac{1}{2\pi}\int_{-\infty}^\infty u(x)e^{-ixz}\,dx \label{eq-5.2B.1} \end{equation} complex $z$?

  1. Observe first that if \begin{equation} |u(x)|\le C\left\{\begin{aligned} &e^{-ax} &&x\le 0,\\ &e^{-bx} &&x\ge 0, \end{aligned}\right. \label{eq-5.2B.2} \end{equation} with $a< b$, then integral (\ref{eq-5.2B.1}) converges in the strip $\{z\colon a<\Im (z) < b\}$ and defines here a holomorphic function, which also tends to $0$ as $|z|\to\infty$, $a+\epsilon \le \Im (z) \le b-\epsilon$ with arbitrary $\epsilon>0$.
  2. In particular, if \begin{align} &u(x)=0 &&\text{for }\ x\le 0 \label{eq-5.2B.3} \end{align} and satisfies (\ref{eq-5.2B.2}), then integral (\ref{eq-5.2B.1}) converges in the lower half-plane $\{z\colon \Im (z) < a\}$ and defines here a holomorphic function, which also tends to $0$ as $|z|\to\infty$, $\Im (z) \le a-\epsilon $ with arbitrary $\epsilon>0$.
  3. On the other hand, (almost) converse is also true due to Paley-Wiener theorem

Paley-Wiener theorem

Theorem 1 (Paley-Wiener theorem). The following statements are equivalent:

  1. $f(z)$ is holomorphic in lower half-plane $\{z\colon \Im (z) < 0\}$, \begin{equation} \int _{-\infty}^\infty |f(\xi + i\eta)|^2\,d\xi \le M\qquad \forall \eta\le 0 \label{eq-5.2B.4} \end{equation}
  2. There exists a function $u$, $u(x)=0$ for $x<0$ and \begin{equation} \int _{-\infty}^\infty |u(x)|^2\,dx\le M \label{eq-5.2B.5} \end{equation} such that $f(z)=\hat{u}(z)$ for $z\colon \Im (z)\le 0$.

To consider functions holomorphic in in the lower half-plane $\{z\colon \Im (z) <a\}$ one needs to apply Paley-Wiener theorem to $g(z)=f(z-ia)$ with $c<a$.

To prove Paley-Wiener theorem one needs to consider Fourier integral \begin{align*} \hat{u}(z)= u(x):=&\int_{-\infty}^{\infty} f(\xi)e^{ix\xi}\,d\xi\\ =&\int_{-\infty+i\eta}^{\infty} f(z)e^{ixz}\,dz \end{align*} where we changed the contour of integration (and one can prove that the integral has not changed) observe, that for $x<0$ this integral tends to $0$ as $\eta\to -\infty$.

Laplace transform

Laplace transform is defined for functions $u: [0,\infty)\to\mathbb{C}$ such that \begin{equation} |u(x)|\le Ce^{ax} \label{eq-5.2B.6} \end{equation} by \begin{equation} \mathcal{L} [u] (p)= \int_0^\infty e^{-px}u(x)\,dx,\qquad \Re (p) >a. \label{eq-5.2B.7} \end{equation} Obviously, it could be described this way: extend $u(x)$ by $0$ to $(-\infty,0)$, then make a Fourier transform (\ref{eq-5.2B.1}), and replace $z=-ip$; then $\Im (z)< a$ translates into $\Re (p)>a$.

Properties of Fourier transform translate into properties Laplace transform, but with a twist \begin{gather} (f*g) (x):=\int_0^x f(y)g(x-y)\,dy, \label{eq-5.2B.8}\\ \mathcal{L}[u'] (p)=p\mathcal{L}[u](p) -pu(0^+). \label{eq-5.2B.9} \end{gather} One can prove (\ref{eq-5.2B.9}}) by integration by parts. Those who are familiar with distributions (see Section 11.1 can obtain it directly because \begin{equation} (\mathcal{E} (u))'(x) = \mathcal{E}(u')(x) +u^+(0) \delta (x), \label{eq-5.2B.10} \end{equation} where $\mathcal{E}$ is an operator of extension by $0$ from $[0,\infty)$ to $(-\infty,\infty)$ and $\delta$ is a Dirac delta function.

The Laplace transform provides a foundation to Operational Calculus by Oliver Heaviside. Its applications to Ordinary Differential Equations could be found in Chapter 6 of Boyce-DiPrima textbook.

Using complex variables

Complex variables could be useful to find Fourier and inverse Fourier transforms of certain functions.

Example 1. Let us find Fourier transform of $\displaystyle{f(x)=\frac{1}{x^2+a^2}}$, $\Re (a) >0$. \begin{equation*} \hat{f}(k)=\frac{1}{2\pi } \int_{-\infty}^\infty \frac{e^{-ikx }\, dx}{x^2+a^2}. \end{equation*} As $k \gtrless 0$ function $\frac{e^{-ikz }}{z^2+a^2}$ is holomorphic at ${z\colon \Im (z) \lessgtr 0}$ except $z=\mp ia $, and nicely decays; then \begin{equation*} \hat{f}(k)=\mp i \operatorname{Res} \bigl( \frac{e^{-ikz }}{z^2+a^2}; z=\mp ia\bigr)= \mp i \frac{e^{-ikz }}{2z}\bigr|_{z=\pm ia}= \frac{1}{2}e^{-|k| }. \end{equation*}

One can apply the same arguments to any rational function $\displaystyle{\frac{P(x)}{Q(x)}}$ where $P(x)$ and $Q(x)$ are polynomials of order $m$ and $n$, $m<n$ and $Q(x)$ does not have real roots.

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