The Fourier Transform

Let’s finally define the Fourier transform, carefully, as usual. In the Fourier series expansion $S_n(f)$ , we have

$\displaystyle (S_n(f))(t) = \sum_{k=-\infty}^{\infty} c_k e^{ikt},\quad c_k := \frac 1{2\pi}\int_{-\pi}^{\pi} f(t) e^{-ikt}\, \mathrm dt.$

If we regard $k$ as a variable, then $k \mapsto c_k$ becomes a function known as the Fourier transform. Denote

$L^1(\mathbb R) := \{f \in \mathcal F(\mathbb R, \mathbb C): f\ \text{is Lebesgue-integrable}\}.$

Definition 1. For any $f \in L^1(\mathbb R)$ , the Fourier transform $\hat f : \mathbb R \to \mathbb C$ is defined by

$\displaystyle \hat f(\xi) := \frac 1{2\pi} \int_{\mathbb R} f(x) e^{-i \xi x}\, \mathrm dx.$

We can extend it to a function $\hat f : \mathbb C \to \mathbb C$ as long as $\mathrm{Im}(\xi) \leq 0$ i.e. if $\xi = \sigma + i \tau$ and $\tau \leq 0$ , then the integral on the right-hand side

$\displaystyle \hat f(\sigma + i \tau) := \frac 1{2\pi} \int_{\mathbb R} f(x) \cdot e^{\tau x} e^{-i \sigma x}\, \mathrm dx$

will still converge as desired.

Remark 1. For any $s \in \mathbb C$ with $\mathrm{Re}(s) \geq 0$ , $\mathrm{Im}(-is) = -\mathrm{Re}(s) \leq 0$ , so that

$\begin{aligned} \hat f(-i s) &= \frac 1{2\pi} \int_{\mathbb R} f(t) e^{- i (-is) t}\, \mathrm dt \\ &= \frac 1{2\pi} \int_{\mathbb R} f(t) e^{-s t}\, \mathrm dt \\ &= \frac 1{2\pi} \cdot \mathcal L \{ f\}(s), \end{aligned}$

where the right-hand side denotes the Laplace transform of $f$ . More compactly,

$\mathcal L \{ f\}(s) = 2\pi \cdot \hat f(-i s),\quad \mathrm{Re}(s) \geq 0.$

In particular, if $\xi \in \mathbb R$ , then $\mathrm{Im}(\xi) = \mathrm{Re}(i\xi) = 0$ , so that the equation

$\displaystyle \hat f(\xi) = \hat f(-i(i\xi)) = \frac 1{2\pi} \cdot \mathcal L\{f\}(i \xi)$

is well-defined.

We make this connection now, so that we can rigorously define the inverse Laplace transform in terms of the inverse Fourier transform, made possible by the Fourier inversion theorem.

Depending on usage or convenience, we denote $\mathcal F \{ f \} = \hat f$ . To emphasise the underlying variables, we denote $\mathcal F\{f(x)\} = \hat f(\xi)$ .

For any subset $K \subseteq \mathbb R$ , denote $f_K = f \cdot \mathbb I_K$ .

Example 1. For piecewise continuously differentiable $2\pi$ -periodic functions $f$ , the Fourier coefficients $c_k$ of the Fourier series of $f$ are given by $c_k = \hat f_{[-\pi, \pi]}(k)$ .

Example 2. For any $a > 0$ , $\hat{\mathbb I}_{[-a,a]}$ is continuous on $\mathbb R$ such that for $\xi \neq 0$ ,

$\displaystyle \hat{\mathbb I}_{[-a, a]}(\xi) = \frac{ \sin(a \xi) }{ \pi\xi }.$

Proof. By the fundamental theorem of calculus, for $\xi \neq 0$ ,

$\begin{aligned}\hat{\mathbb I}_{[-a, a]}(\xi) &= \frac 1{2\pi} \int_{\mathbb R} \mathbb I_{[-a, a]}(x) \cdot e^{-i \xi x}\, \mathrm dx \\ &= \frac 1{2\pi} \int_{-a}^{a} e^{-i \xi x}\, \mathrm dx \\ &= \frac 1{2\pi} \cdot \frac{e^{-i a\xi} - e^{i a\xi}}{- i \xi} \\ &= \frac 1{ \pi\xi } \cdot \frac{e^{i a\xi} - e^{-i a\xi}}{ 2i} = \frac{\sin(a \xi)}{\pi \xi}. \end{aligned}$

By the usual sine limit,

$\begin{aligned} \lim_{\xi \to 0}\hat{\mathbb I}_{[-a, a]}(\xi) &= \frac {a}{\pi} \cdot \lim_{\xi \to 0} \frac{\sin(a \xi)}{a \xi} = \frac{a}{\pi} \cdot 1 \\ &= \frac{a}{\pi} = \frac 1{2\pi} \int_{-\infty}^{\infty} \mathbb I_{[-a, a]}(x) \, \mathrm dx = \hat{\mathbb I}_{[-a, a]}(0), \end{aligned}$

establishing continuity.

Example 3. $\mathcal F\{x f(x)\} = i\hat f'(\xi)$ .

Proof. Differentiating under the integral sign,

$\begin{aligned} \mathcal F\{x f(x)\} &= \frac 1{2\pi} \int_{\mathbb R} xf(x) e^{-i \xi x}\, \mathrm dx \\ &= \frac 1{2\pi} \int_{\mathbb R} f(x) \cdot i \cdot \frac{\partial}{\partial \xi} (e^{-i \xi x})\, \mathrm dx \\ &= i \cdot \frac{\mathrm d}{\mathrm d\xi} \left( \frac 1{2\pi} \int_{\mathbb R} f(x) e^{-i \xi x}\, \mathrm dx \right) = i \hat f'(\xi). \end{aligned}$

Theorem 1. For any $a > 0$ , $\displaystyle \mathcal F\{ e^{-ax^2}\} = \frac 1{2\sqrt{a\pi}} \cdot e^{-\frac{\xi^2}{4a}}$ .

Proof. We need to evaluate the integral

$\displaystyle \mathcal F\{ e^{-ax^2}\} = \frac 1{2\pi} \int_{\mathbb R} e^{-ax^2} e^{-i \xi x}\, \mathrm dx = \frac 1{2\pi} \cdot \lim_{R \to \infty} \int_{-R}^R e^{-(ax^2+i \xi x)}\, \mathrm dx.$

Completing the square,

$\displaystyle ax^2 + i\xi x = a\left( x + \frac{i \xi}{2a} \right)^2 + \frac{\xi^2}{4a},$

so that

$\displaystyle \mathcal F\{ e^{-ax^2}\} =\frac 1{2\pi} \cdot e^{-\frac{\xi^2}{4a}} \cdot \lim_{R \to \infty} \int_{-R}^R e^{-a\left( x + \frac{i \xi}{2a} \right)^2 }\, \mathrm dx.$

To evaluate the right-hand side, we return to contour integration. Define the entire function $g(z) := e^{-az^2}$ . Consider the anti-clockwise rectangle

$\Gamma : -R \to R \to R + i\xi/2a \to -R + i\xi/2a.$

Denote the corresponding by $\Gamma_1,\Gamma_2,\Gamma_3,\Gamma_4$ respectively. By the Cauchy-Goursat theorem,

$\displaystyle \oint_{\Gamma} g(z)\, \mathrm dz = 0.$

We parameterise the integrals over the four sides:

$\begin{aligned} \oint_{\Gamma_1} g(z)\, \mathrm dz &= \int_{-R}^{R} g(t)\, \mathrm dt \\ \oint_{\Gamma_2} g(z)\, \mathrm dz &= i\int_{0}^{ \xi/2a} g(R + it)\, \mathrm dt, \\ \oint_{\Gamma_3} g(z)\, \mathrm dz &= -\int_{-R}^{R} g(t + i \xi/2a)\, \mathrm dt, \\ \oint_{\Gamma_4} g(z)\, \mathrm dz &= -i\int_0^{\xi/2a} g(-R + it )\, \mathrm dt, \end{aligned}$

so that

$\displaystyle \left( \oint_{\Gamma_1} + \oint_{\Gamma_3}\right) g(z)\, \mathrm dz = - \left( \oint_{\Gamma_2} + \oint_{\Gamma_4}\right) g(z)\, \mathrm dz.$

On $\Gamma_2$ ,

$|g(R+it)| = |e^{-a(R+it)^2}| = e^{-a(R^2 - t^2)} \leq e^{\xi^2/4a^2} \cdot e^{-aR^2}.$

Similarly, on $\Gamma_4$ , $|g(-R+it)| \leq e^{\xi^2/4a^2} \cdot e^{-aR^2}$ . By the ML-inequality,

$\begin{aligned}\left| \left( \oint_{\Gamma_2} + \oint_{\Gamma_4}\right) g(z) \right| &\leq 2 \cdot e^{\xi^2/4a^2} \cdot e^{-aR^2}. \end{aligned}$

Therefore,

$\begin{aligned} \left| \oint_{\Gamma_3} g(z)\, \mathrm dz - \oint_{\Gamma_1} g(z)\, \mathrm dz \right| &= \left| \left( \oint_{\Gamma_2} + \oint_{\Gamma_4}\right) g(z)\, \mathrm dz \right| \leq 2 \cdot e^{\xi^2/4a^2} \cdot e^{-aR^2}. \end{aligned}$

Taking $R \to \infty$ , since the Gaussian integral yields

$\displaystyle \lim_{R \to \infty} \oint_{\Gamma_1} g(z)\, \mathrm dz = \lim_{R \to \infty} \int_{-R}^{R} e^{-at^2}\, \mathrm dt = \sqrt{\pi/a},$

we have

$\displaystyle \lim_{R \to \infty} \oint_{\Gamma_3} g(z)\, \mathrm dz = \lim_{R \to \infty} \oint_{\Gamma_1} g(z)\, \mathrm dz = \sqrt{\pi/a}.$

Therefore,

$\begin{aligned} \lim_{R \to \infty} \int_{-R}^{R} e^{-a(x + i \xi/2a)^2}\, \mathrm dx &= \lim_{R \to \infty} \int_{-R}^{R} g(x + i \xi/2a)\, \mathrm dx \\ &= \lim_{R \to \infty} \oint_{\Gamma_3} g(z)\, \mathrm dz = \sqrt{\pi/a}. \end{aligned}$

Finally,

$\displaystyle \mathcal F\{ e^{-ax^2}\} = \frac 1{2\pi} \cdot e^{-\frac{\xi^2}{4a}} \cdot \sqrt{\frac{\pi}{a}} = \frac 1{2\sqrt{a\pi}} \cdot e^{-\frac{\xi^2}{4a}}.$

Remark 1. Setting $a = 1/2$ , we obtain $\displaystyle \mathcal F\{ e^{-x^2/2}\} = \frac 1{\sqrt{2\pi}} \cdot e^{-\xi^2/2}$ .

We explore inverting the Fourier transform in the next post, in order to finally establish the injectivity of Laplace transforms.

—Joel Kindiak, 26 Aug 25, 1920H

KindiakMath

The Fourier Transform

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The Fourier Transform

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