5.2B. Fourier transform in the complex domain

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### Introduction

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 statents 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\ge 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 $z\colon \Im z\ge 0$.

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

### 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.