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Introduction to Linear Algebra with Mathematica

Glossary

Preface

This section is about a classical integral transformation, known as the Fourier transformation. This is the most important and most personal discovery of Fourier, in which he had no predecessors. Since the Fourier transform is expressed through an indefinite integral, its numerical evaluation is an ill-posed problem. It is a custom to use the Cauchy principal value regularization for its definition, as well as for its inverse. Fourier transform gives the spectral decomposition of the derivative operator (more precisely, the momentum operator) \( {\bf j}\,\texttt{D} , \) where \( \texttt{D} = {\text d}/{\text d}x \) and j is the unit vector in the positive vertical direction on the complex plane ℂ. Fourier transformation and its applications is a wide area and books are written for this subject. Therefore, we are forced to include only basic results that we cannot avoid when dealing with differential equations.

Fourier transform

As we have seen in the previous chapter, Fourier series is a useful tool for representing either periodic functions or functions confined in limited range of interest. However, in many problems, the function of interests, such as a single unrepeated pulse of force or voltage, is nonperiodic over an infinite range. In such a case, we can still imagine that the function is periodic with the period approaching infinity. In this limit, the Fourier series becomes the Fourier integral. To extend the concept of Fourier series to nonperiodic functions, let us consider an absolutely integable function f ∈ 𝔏¹(ℝ) that satisfies Dirichlet consitions at every finite subinterval.

A standard Fourier series on the circle 𝕋 = ℝ/(2ℓℕ) of length 2ℓ, −ℓ ≤ t ≤ ℓ, can be written either in exponential form

\[ f(t) \sim \mbox{P.V.} \sum_{n=-\infty}^{+\infty} \hat{f}(n)\, e^{{\bf j}n\pi t/\ell} = S[f](t) \]
or in trigonometric form
\[ f(t) \sim \frac{a_0}{2} + \sum_{k\ge 1} \left[ a_k \cos \left( \frac{\pi}{\ell}\,k t \right) + b_k \sin \left( \frac{\pi}{\ell}\, k t \right) \right] = S[f] . \]
Here j is a unite imaginary vector on the complex plane ℂ, so j² = −1, and «P.V.» is the abbreviation for Cauchy principal value regularization. The series S[f] may be thought of as an expansion of a periodic function (of period T = 2ℓ) into simple harmonics. The coefficients in the Fourier expansion are determined via Euler--Fourier formulas:
\[ \hat{f} (n) = \frac{1}{2\ell} \int_{-\ell}^{\ell} f(x)\,e^{-n{\bf j}\pi x/\ell} {\text d} x, \qquad n \in \mathbb{Z} = \left\{ 0, \pm 1, \pm 2, \ldots \right\} , \]
or
\[ \begin{bmatrix} a_k (f) \\ b_k (f) \end{bmatrix} = \frac{1}{\ell} \int_{-\ell}^{\ell} f(x) \begin{bmatrix} \cos \left( \frac{\pi}{\ell}\,k x \right) \\ \sin \left( \frac{\pi}{\ell}\, k x \right) \end{bmatrix} {\text d} x , \qquad k \in \mathbb{N} = \left\{ 0, 1, 2, 3, \ldots \right\} . \]
Clearly, the choice of period T and the way of considering the Fourier series either in exponential or trigonometric form is just a matter of convenience because the finite partial sums are the same:
\begin{align*} S_N (f; x) &= \sum_{n=-N}^{N} \hat{f} (n)\, e^{n{\bf j}\pi x/\ell} \\ &= \frac{a_0}{2} + \sum_{k=1}^N \left[ a_k \cos \left( \frac{\pi}{\ell}\, k x \right) + b_k \sin \left( \frac{\pi}{\ell}\, k x \right) \right] \\ &= \frac{1}{\ell} \int_{-\ell}^{\ell} f(y)\, D_N \left( \frac{\pi \left( x-y \right)}{\ell} \right) = \frac{1}{\ell} \left( f \star D_N \right) (x) , \end{align*}
where the Dirichlet kernel is
\[ D_N (x) = \frac{1}{2} \sum_{n=-N}^N e^{n{\bf j}\pi x/\ell} = \frac{1}{2} + \sum_{k=1}^N \cos kz = \frac{\sin \left( N + \frac{1}{2} \right) z}{2\,\sin \frac{z}{2}} , \quad z = \frac{n\pi}{\ell} . \]
Each individual term either \( \displaystyle e^{n{\bf j}\pi t/\ell} \) or \( \displaystyle \cos \frac{k\pi t}{\ell} , \quad \sin \frac{k\pi t}{\ell} \) in Fourier series is a periodic function. Its period Tk is determined by the relation that when t is increased by Tk, the function returns to its previous value,
\[ \sin \frac{k\pi}{\ell} \left( t + T_k \right) = \sin \left( \frac{k\pi}{\ell}\, t + \frac{k\pi}{\ell}\, T_k \right) = \sin \frac{k\pi}{\ell}\, t . \]
Thus,
\[ \frac{k\pi}{\ell}\, T_k = 2\pi \qquad \Longrightarrow \qquad T_k = \frac{2\ell}{k} , \quad k \ne 0. \]
The (temporal) frequency ν is defined as the number of oscillations in one second (unit time). Therefore, each term is associated with a frequency νn:
\[ \nu_n = \frac{1}{T_n} = \frac{n}{2\ell} , \qquad n=1,2,\ldots . \]
Now if t stands for time, then νn is just the usual temporal frequency. If the variable is x, standing for distance, νn is simply the spatial frequency. The distribution of the set of all frequencies νn is called the frequency spectrum (plural: frequency spectra). To see what happens to the frequency spectra as ℓ increases, consider the cases where ℓ = 1, 5, 10, and 100. The corresponding frequencies of the spectra are as follows:

            n = 0       n = 1       n = 2       n = 3       n = 4       …
ℓ = 1,       νn =       0.0 ,       0.50       1.0       1.5       2.0       …
ℓ = 5,       νn =       0.0,       0.10       0.2       0.3       0.4       …
ℓ = 10,       νn =       0.0 ,       0.05       0.10       0.15       0.20       …
ℓ = 100,       νn =       0.0 ,       0.005       0.10       0.015       0.020       …

It is seen that as ℓ increases, the discrete spectrum becomes more and more dense. It will approach a continuous spectrum as ℓ → ∞, and the Fourier series appears to be an integral. This is indeed the case, if f(t) is absolutely integrable over the infinite range.

Often the angular frequency, defined as ωn = 2π νn, is used to simplify the writing. Since

\[ \omega_n = 2\pi\,\nu_n = 2\pi \,\frac{n}{2\ell} = \frac{n\pi}{\ell} \]
the Fourier series can be written as
\[ S[f](t) = \frac{1}{2\ell} \int_{-\ell}^{\ell} f(x)\,{\text d} x + \sum_{k\ge 1} \left[ a_k \cos \omega_k t + b_k \sin \omega_k t \right] = \sum_{n=-\infty}^{\infty} \hat{f}(n) \, e^{n{\bf j}\pi t/\ell} . \]
According to our assumption that f(t) is absolutely integrable over the infinite range, this means that f ∈ 𝔏¹(ℝ) even when ℓ → ∞. Therefore,
\[ \lim_{\ell \to \infty} \frac{1}{2\ell} \int_{-\ell}^{\ell} f(x)\,{\text d} x = 0 . \]
Hence,
\[ S[f](t) = \sum_{k\ge 1} \left[ a_k \cos \omega_k t + b_k \sin \omega_k t \right] = \sum_{n\ne 0} \hat{f}(n) \, e^{n{\bf j}\pi t/\ell} . \]
where
\[ \begin{bmatrix} a_k \\ b_k \end{bmatrix} = \frac{1}{\ell} \int_{-\ell}^{\ell} f(x) \begin{bmatrix} \cos\omega_k x \\ \sin\omega_k x \end{bmatrix} {\text d} x , \qquad \hat{f}(n) = \frac{1}{2\ell} \int_{-\ell}^{\ell} f(x) \,e^{n{\bf j}\pi x/\ell} {\text d} x. \]
Furthermore, we can define
\[ \Delta \omega = \omega_{n+1} - \omega_n = \frac{(n+1) \pi}{\ell} - \frac{n\pi}{\ell} = \frac{\pi}{\ell} . \]
So
\[ S[f] = \sum_{k\ge 1} \left[ \frac{\Delta \omega}{\pi} \int_{-\ell}^{\ell} f(x)\,\cos\omega_k x \,{\text d} x \right] \cos\omega_k t + \sum_{k\ge 1} \left[ \frac{\Delta \omega}{\pi} \int_{-\ell}^{\ell} f(x)\,\sin\omega_k x \,{\text d} x \right] \sin\omega_k t , \]
or
\[ S[f] = \sum_{n\ne 0} \frac{\Delta\omega}{2\pi} \int_{-\ell}^{\ell} f(x)\,e^{-{\bf j} \omega_n x} {\text d} x \, e^{{\bf j} \omega_n t} . \]
If we write the series as
\[ S[f] = \Delta\omega \,\sum_{k\ge 1} \left[ A_{\ell}(\omega_k )\,\cos\omega_k t + B_{\ell} (\omega_k )\,\sin\omega_k t \right] = \Delta\omega \,\sum_{n\ne 0} C_{\ell}(\omega_n )\,e^{{\bf j}\omega_n t} , \]
then
\begin{align*} C_{\ell}(\omega ) &= \frac{1}{2\pi} \int_{-\ell}^{\ell} f(x)\, e^{-{\bf j} \omega x} {\text d} x , \\ A_{\ell}(\omega ) &= \frac{1}{\pi} \int_{-\ell}^{\ell} f(x)\,\cos\omega x\,{\text d} x , \qquad B_{\ell}(\omega ) = \frac{1}{\pi} \int_{-\ell}^{\ell} f(x)\,\sin\omega x\,{\text d} x . \end{align*}
Now if we let ℓ → ∞, then ∆ω → 0 and ωn becomes a continuous variable. Furthermore, let
\begin{align*} C(\omega ) &= \lim_{\ell\to\infty} C_{\ell} (\omega_n ) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} f(x)\, e^{{\bf j}\omega x} {\text d} x , \\ A(\omega ) &= \lim_{\ell\to\infty} A_{\ell} (\omega_k ) = \frac{1}{\pi} \int_{-\infty}^{+\infty} f(x)\,\cos \omega x \,{\text d} x , \\ B(\omega ) &= \lim_{\ell\to\infty} B_{\ell} (\omega_k ) = \frac{1}{\pi} \int_{-\infty}^{+\infty} f(x)\,\sin \omega x \,{\text d} x . \end{align*}
Since S[f] remains the same and for any ℓ gives function f, the infinite series becomes a Riemann sum of an integral as ℓ → ∞:
\begin{align*} f(t) &= \int_{-\infty}^{+\infty} C(\omega )\,e^{{\bf j}\omega t} {\text d} \omega , \\ f(t) &= \int_0^{+\infty} \left[ A(\omega )\,\cos\omega t + B(\omega )\,\sin\omega t \right] {\text d} \omega . \end{align*}
Each of these integrals is known as Fourier integral. This development is purely formal. However, it can be made rigorous, which is shown in the following example.

Example 1: We assume that function f() is absolutely integable and satisfies Dirichlet consitions at every finite subinterval, so
\[ \int_{-\infty}^{+\infty} |f(x)|\,{\text d}x < \infty . \]
Then according to Dirichelet's theorem, we have
\[ f(x) = \frac{a_0}{2} + \sum_{k\ge 1} \left[ a_k \cos \left( \frac{\pi}{\ell}\, k x \right) + b_k \sin \left( \frac{\pi}{\ell}\, k x \right) \right] \]
at every finite interval (−ℓ, ℓ) assuming that function f is continuous at x. For a periodic function f of period T, its Fourier coefficients are
\[ \hat{f}(n) = \frac{1}{T} \int_{-T/2}^{T/2} f(x) \,e^{-2\pi{\bf j} nx/T} {\text d} x = \frac{\Delta \omega}{2\pi} \int_{-T/2}^{T/2} f(x) \,e^{-{\bf j} \omega_n x} {\text d} x , \]
where Δω = 2π/T and ωn = nΔω. The integrand is a periodic function,
\[ f(x+T) = f(x) \qquad \mbox{and} \qquad e^{2\pi{\bf j}n} = 1. \]
Therefore,
\[ \hat{f}(n) = \frac{1}{T} \int_{-T/2}^{T/2} f(x) \,e^{-2\pi{\bf j} nx/T} {\text d} x = \int_{-nT/2}^{nT/2} f(x) \,e^{-2\pi{\bf j} nx/T} {\text d} x = \int_{-nT/2}^{nT/2} f(x) \,e^{-{\bf j}\,\omega_n x} = \hat{f}\left( \omega_n \right) . \]
End of Example 1

There are several common conventions for defining the Fourier transform of an integrable complex-valued function f : ℝ → ℂ. In applications, the function f(x) is usually referred to as a signal. Here we will use the following definition, which is most common in applications. The Fourier transform of the function f is traditionally denoted by adding a circumflex: \( \displaystyle {\hat {f}} \) or \( ℱ\left[ f \right] \) or \( f^F . \) Actually, the Fourier transform measures the frequency content of the signal f.

The exponential Fourier transform (or spectrum) of the function f is the complex-valued function defined for the real variable ξ defined (if it exists) as an improper Riemann integral
\begin{equation} \label{EqT.1} \hat{f} (\xi ) =ℱ\left[ f(x) \right] (\xi ) = f^F (\xi ) = \int_{-\infty}^{\infty} f(t)\,e^{{\bf j} \xi\cdot t} \,{\text d}t \end{equation}
with the inverse (that is valid for functions \( \displaystyle {\hat {f}} \) satisfying the Dirichlet conditions)
\begin{align} ℱ^{-1} \left( \hat{f} \right) &= \text{P.V.} \frac{1}{2\pi} \int_{-\infty}^{\infty} \hat{f}(\xi )\,e^{-{\bf j} \xi\cdot t} \,{\text d}\xi = \lim_{N\to \infty} \frac{1}{2\pi} \int_{-N}^N \hat{f}(\xi )\,e^{-{\bf j} \xi\cdot t} \,{\text d}\xi \notag \\ &= \frac{f(t+0) + f(t-0)}{2} . \label{EqT.2} \end{align}
This definition can be extended for the multidimensional case:
\begin{equation} \label{EqT.3} \hat{f} (\xi ) = ℱ(f) = f^F = \int_{{\mathbb R}^n} f(t)\,e^{{\bf j} \xi\cdot t} \,{\text d}t, \quad f(t) = ℱ^{-1} \left( \hat{f} \right) = \frac{1}{(2\pi )^n} \int_{{\mathbb R}^n} \hat{f}(\xi )\,e^{-{\bf j} \xi\cdot t} \,{\text d}\xi , \qquad \end{equation}
where \( \xi\cdot t = \xi_1 t_1 + \xi_2 t_2 + \cdots + \xi_n t_n \) is the inner product and j is the unit vector in the positive vertical direction on the complex plane ℂ so j² = -1. The prefix V.P. indicates that the improper integral is evaluated in the Cauchy principal value sense. The functions f and \( \displaystyle {\hat {f}} \) often are referred to as a Fourier integral pair or Fourier transform pair.
The units of variable ξ in Fourier transform formula \eqref{EqT.1} should be reciprocal to variable t because their product must be dimensionless. Indeed, expanding exponential function into Maclaurin power series \( \displaystyle e^u = 1 + u + \frac{u^2}{2} + \frac{u^3}{3!} + \cdots , \) we see that all powers of u = tξ should have the same dimension, which requires u to be dimensionless. If t has time units, then ξ should have units of frequency in order to make their product dimensionless.

The Fourier transform is usually defined in some function spaces. The most common spaces are defined below.

Let 𝔏¹(ℝ) denote the space of functions on ℝ that are Lebesgue-integrable on (−∞, ∞) functions with the norm
\[ \| f \| = \| f \|_1 = \int_{-\infty}^{\infty} \left\vert f(t) \right\vert {\text d}t < \infty . \]
In this space, all functions which are equal almost everywhere are identified.
The space 𝔏∞(ℝ) is the vector space of bounded measurable functions on ℝ, with the supremum norm (also called the uniform norm or the Chebyshev norm or the infinity norm) defined by
\[ \| f \|_{\infty} = \sup_{x\in \mathbb{R}} \left\vert f(x) \right\vert . \]
Under this norm, the space of all continuous functions is a Banach space that is denoted as ℭ(ℝ) or C(ℝ).
Theorem 1: Let f∈𝔏¹(ℝ) be a function which is Lebesgue-integrable on ℝ. Then its Fourier transform ξ → fF(ξ) of f is a continuous bounded function for all ξ ∈ ℝ. So \( ℱ \) : 𝔏¹(ℝ) → ℭ(ℝ) ⊂ 𝔏∞(ℝ) and \( \| f^{F} \|_{\infty} \le \| f \|_1 . \)
Observe that for any \( \xi \in {\mathbb R}^n ,\)
\[ \left\vert \hat{f} \right\vert \le \int | f (t) |\,{\text d}t = \| f \|_1 . \]
Therefore, the integral of Fourier transform of f∈𝔏¹(ℝ) converges uniformly with respect to ξ on ℝ.

To show that Fourier transform fF is continuous, let 𝑎 and b be real numbers with 𝑎 < b and use the triangle inequality to show that

\begin{align*} \left\vert \hat{f}(\xi ) - \hat{f}(\eta ) \right\vert &\leqslant \left\vert \int_{-\infty}^{+\infty} f(x)\,e^{{\bf j}x\xi} {\text d} x - \int_{a}^{b} f(x)\,e^{{\bf j}x\xi} {\text d} x \right\vert \\ & \quad + \left\vert \int_{-\infty}^{+\infty} f(x)\,e^{{\bf j}x\eta} {\text d} x - \int_{a}^{b} f(x)\,e^{{\bf j}x\eta} {\text d} x \right\vert \\ & \quad + \left\vert \int_{a}^{b} f(x)\,e^{{\bf j}x\xi} {\text d} x - \int_{a}^{b} f(x)\,e^{{\bf j}x\eta} {\text d} x \right\vert . \end{align*}
By the uniform convergence of the integral defining fF, the numbers 𝑎 and b can be chosen so that the first two terms on the fight-hand side above are as small as desired. With 𝑎 and b already chosen, the third term, which is no larger than
\[ \max_{x \in [a, b]} \left\vert e^{{\bf j}x\xi} - e^{{\bf j}x\eta} \right\vert \int_{-\infty}^{+\infty} \left\vert f(x) \right\vert {\text d} x , \]
can also be made as small as desired by the continuity of the exponential if ξ − η is small enough. This shows that Fourier transform fF is continuous.    ◂
Example 2: The Fourier transform of the lorentzian function with parameter 𝑎 > 0
\[ f(x) = \frac{a}{a^2 + x^2} \]
is
\[ f^F (\xi ) = ℱ \left[ \frac{a}{a^2 + x^2} \right] = \pi\,e^{-a|\xi |} \]
because
Assuming[a > 0 && t > 0, Integrate[a/(x^2 + a^2)*Exp[I*x*t], {x, -Infinity, Infinity}]]
E^(-a t) \[Pi]
The corresponding integral evaluation is based on Cauchy residue theorem and Jordan's lemma. For ξ > 0, we have
\[ f^F (\xi ) = 2\pi{\bf j}\,\mbox{Res}_{x = {\bf j}a} \,\frac{a}{a^2 + x^2}\, e^{{\bf j}x\xi} = 2\pi{\bf j}\,\lim_{x\to {\bf j}a} \,\frac{a}{2x} \, e^{{\bf j}x\xi} = \pi\,e^{-a|\xi |} . \]
For ξ < 0, we need to evaluate the residue at x = −j𝑎:
\[ f^F (\xi ) = -2\pi{\bf j}\,\mbox{Res}_{x = -{\bf j}a} \,\frac{a}{a^2 + x^2}\, e^{{\bf j}x\xi} = -2\pi{\bf j}\,\lim_{x\to -{\bf j}a} \,\frac{a}{2x} \, e^{{\bf j}x\xi} = \pi\,e^{-a|\xi |} . \]
   ■
Theorem 2: Let f ∈ 𝔏¹(ℝ) be an integrable function such that its Fourier transform fF is itself integrable. Then
\[ ℱ^{-1} \left[ f^F \right] = f(x) \qquad \mbox{for almost all }x \in\mathbb{R} . \]
There are some functions that are not necessarily integrable, but whose square is. Such is the “sine cardinal” function: \( \mbox{sinc}(x) = x^{-1} \sin x . \) It turns out that in physics, square integrable functions are of paramount importance and occur frequently. It is therefore advisable to extend the definition of the Fourier transform to the class of square integrable functions.
Let 𝔏²(ℝ) denote the space of measurable functions defined on ℝ, with complex values (up to equality almost everywhere) which are square integrable, with the norm
\[ \| f \|_2 = \left( \int_{-\infty}^{\infty} \left\vert f(t) \right\vert^2 {\text d}t \right)^{1/2} < \infty . \]
With the inner product
\[ \left\langle f , g \right\rangle = \int_{-\infty}^{\infty} \overline{f}(t)\,g(t)\, {\text d}t = \left\langle f \.\vert\, g \right\rangle , \]
𝔏²(ℝ) becomes the Hilbert space. The Fourier transform maps 𝔏²(ℝ) → 𝔏²(ℝ) as an (isometric) isomorphism.

A fundamental result due to Riesz and Fischer, which is not obvious at all, shows that 𝔏²(ℝ) is complete.

Note that the inverse Fourier transformation is an ill-posed problem; therefore, its application must be done with care (see Kabanikhin's survey). In the above definition, the principal value means that the limit is taken over symmetrical interval

\begin{equation} \label{EqT.4} \text{P.V.} \int_{-\infty}^{\infty} \hat{f}(\xi )\,e^{-{\bf j} \xi\cdot t} \,{\text d}\xi = \lim_{N\to \infty} \, \int_{-N}^N \hat{f}(\xi )\,e^{-{\bf j} \xi\cdot t} \,{\text d}\xi \end{equation}
while ordinary definition requires their independence:
\[ \int_{-\infty}^{\infty} \hat{f}(\xi )\,e^{-{\bf j} \xi\cdot t} \,{\text d}\xi = \lim_{N, M\to \infty} \, \int_{-N}^M \hat{f}(\xi )\,e^{-{\bf j} \xi\cdot t} \,{\text d}\xi . \]
So the standard definition requires that both bounds N and M approach infinity independently of each other. The Fourier transformation and its inverse are a bounded operations in the space of square integrable functions denoted by 𝔏² (other common notations are L² or L2). The following equality was proved hy Michel Plancherel in 1910. It is also known as Parseval's formula:
Plancherel's Theorem: The Fourier transform is an isometry on 𝔏², that is, \( \| f \|_{𝔏^2}^2 = \int_{-\infty}^{\infty} | f(x) |^2 {\text d}x = \left( 2\pi \right)^{-1} \int_{-\infty}^{\infty} | \hat{f}(\xi ) |^2 {\text d}\xi = \left( 2\pi \right)^{-1} \| \hat{f} \|_{𝔏^2}^2 . \) In multidimensional case, we have
\begin{equation} \label{EqT.5} \| \hat{f} \|_2^2 = \int_{-\infty}^{\infty} \left\vert \hat{f}(\xi ) \right\vert^2 {\text d}\xi = 2\pi\, \int_{-\infty}^{\infty} \left\vert f(x ) \right\vert^2 {\text d}x = 2\pi\, \| f \|^2 . \end{equation}

The Plancherel's identity can be extended for n-dimensional case:
\[ \langle \hat{f}, \hat{g} \rangle = \left( 2\pi \right)^n \langle f, g \rangle . \]

Recall that a real-valued function f : ℝ → ℝ is called an absolutely integrable function if it is a function whose absolute value is integrable, meaning that the integral of the absolute value over the whole domain is finite:

\begin{equation} \label{EqT.6} \| f \|_1 = \int_{-\infty}^{\infty} \left\vert f(x) \right\vert {\text d} x < \infty . \end{equation}
We abbreviate it as f ∈ 𝔏¹(ℝ). A square-integrable function, also called a quadratically integrable function or 𝔏²(ℝ) is a real- or complex-valued measurable function for which the integral of the square of the absolute value is finite:
\begin{equation} \label{EqT.7} \| f \|^2_2 = \int_{-\infty}^{\infty} \left\vert f(x) \right\vert^2 {\text d} x < \infty . \end{equation}
A function of bounded variation is a real-valued function whose total variation is bounded (finite). This is equivalent to say that the function has on a compact interval finite number of maximum and minimum; a function of finite variation can be represented by the difference of two monotonic functions having discontinuities, but at most countably many. Obviously, a function of bounded variation cannot have an infinite jump. A real-valued function f is said to satisfy the Dirichlet conditions if A sufficient condition for existence of the Fourier transform \( \displaystyle {\hat {f}} \) is its absolutely integrability, f ∈ 𝔏¹(ℝ). In this case, \( \displaystyle {\hat {f}} \) is uniformly continuous on ℝ and
\begin{equation} \label{EqT.8} \lim_{|\xi | \to \infty} \hat {f} (\xi ) = 0 . \end{equation}

The inversion Fourier transformation formula is based on the following statement.

Lemma: If a function f(x+u) satisfies the Dirichlet conditions in an interval 𝑎 < u < b, then
\[ \lim_{\omega \to \infty} \frac{2}{\pi} \int_a^b f(x+u) \,\frac{\sin\,\omega u}{\omega}\,{\text d} u = \begin{cases} f(x+0) + f(x-0) , & \ \mbox{ if } a < 0 < b , \\ f(x+0) , & \ \mbox{ if } a=0 < b , \\ f(x-0) , & \ \mbox{ if } a < 0 = b , \\ 0 , & \ 0 < a < b \mbox{ or } a < b < 0 . \end{cases} \]
The Fourier integral exists not for arbitrary functions, but for functions that approached zero at infinity. Although the exponential Fourier integral may not converge for a particular function f, the following functions
\[ \begin{split} \hat{f}_{+} (\xi ) &= \int_0^{\infty} f(t)\, e^{{\bf j}\xi t}\,{\text d}t , \\ \hat{f}_{-} (\xi ) &= \int_{-\infty}^0 f(t)\, e^{{\bf j}\xi t}\,{\text d}t , \end{split} \]
where ξ = u + jv, may exist: the latter for large enough positive v, and the former for large in absolute value negative v. The inverse Fourier transform gives
\[ \frac{1}{2\pi} \int_{-\infty}^{\infty} \hat{f}_{+} (u+{\bf j}v ) \, e^{-{\bf j}(u + {\bf j}v)x} \,{\text d}u = \begin{cases} f(x) , & \ x> 0, \\ 0, & \ x < 0. \end{cases} \]

The statement that f can be reconstructed from \( \displaystyle {\hat {f}} \) is known as the Fourier inversion theorem, and was first introduced in Fourier's Analytical Theory of Heat.

Fourier integral Theorem: If a piecewise continuous function f(x) is of finite variation and absolutely integrable on (-∞,∞), then
\[ \frac{f(x+0) + f(x-0)}{2} = \frac{1}{\pi} \int_0^{\infty} {\text d}s \mbox{ V.P.}\int_{-\infty}^{\infty} {\text d}t\, f(t)\,\cos s(t-x) . \]
In other words, the above equation holds for functions satisfying the Dirichlet conditions.
Let f(x) be a 2ℓ-periodic function of finite variation that can be represented by Fourier series:
\[ \frac{f(x+0) + f(x-0)}{2} = \frac{a_0}{2} + \sum_{n\ge 1} \left[ a_n \cos \frac{n\pi\,x}{\ell} + b_n \sin \frac{n\pi\,x}{\ell} \right] . \]
The coefficients are defined by the Euler--Fourier formulas:
\[ a_n = \frac{1}{\ell} \int_{-\ell}^{\ell} f(t)\,\cos \frac{n\pi\,t}{\ell} \,{\text d}t , \qquad b_n = \frac{1}{\ell} \int_{-\ell}^{\ell} f(t)\,\sin \frac{n\pi\,t}{\ell} \,{\text d}t . \]
Substituting these values into the Fourier series, we obtain
\[ \frac{f(x+0) + f(x-0)}{2} = \frac{1}{2\ell} \int_{-\ell}^{\ell} f(t)\,{\text d}t + \sum_{n\ge 1} \frac{1}{\ell} \int_{-\ell}^{\ell} f(t)\,\cos \frac{n\pi \left( x-t \right)}{\ell}\,{\text d}t \]
If we set u = nπ/ℓ, δu = 1/ℓ and find the formal limit when ℓ → ∞, then the infinite series becomes the integral and it leads to
\[ \frac{f(x+0) + f(x-0)}{2} = \frac{1}{\pi} \int_0^{\infty} {\text d}u \int_{-\infty}^{\infty} f(t)\,\cos u\left( x-t \right) {\text d}t . \]
   ◂
Based on Euler's formula \( \cos t = \frac{1}{2} \left( e^{{\bf j}t} + e^{-{\bf j}t} \right) , \) the right-hand side can be rewritten as
\begin{align*} f(x) &= \frac{1}{\pi} \int_0^{\infty} {\text d}u \int_{-\infty}^{\infty} f(t)\,\cos u\left( x-t \right) {\text d}t = \frac{1}{2\pi} \int_0^{\infty} {\text d}u \int_{-\infty}^{\infty} f(t)\left[ e^{{\bf j} (x-t)u} + e^{-{\bf j} (x-t)u} \right] {\text d}t \\ &= \frac{1}{2\pi} \int_{-\infty}^{\infty} {\text d}u \int_{-\infty}^{\infty} f(t)\, e^{{\bf j} (x-t)u} {\text d}t = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{{\bf j} xu} \,{\text d}u \int_{-\infty}^{\infty} f(t)\, e^{-{\bf j} tu} {\text d}t . \end{align*}
Theorem 4: Let f be an absolutely continuous function on ℝ such that f and its derivative f′ are both in 𝔏¹(ℝ). For each x ∈ ℝ,
\[ f(x) = \frac{1}{2\pi}\,\lim_{N\to\infty} \int_{-N}^N {\text d}\xi\,e^{-{\bf j}x\xi} \int_{-\infty}^{\infty} {\text d}t\,e^{{\bf j}t\xi} f(t) , \qquad {\bf j}^2 =-1. \]
Example 3: Suppose a function f(t) and its derivative df/dt are both piecewise continuous functions for all t ≥ 0. If the function f(t) is not integrable, then its Fourier transform does not exist. However, it may happen that f(t) times an exponentially decay function could be integrable, so the Fourier transform of such product exists. We define an extension
\[ g(t) = \begin{cases} e^{-ct} f(t) , & \ \mbox{ for } t \ge 0, \\ 0, & \ \mbox{ for } t < 0, \end{cases} \]
where c is a positive real constant. Then obviously \( \int_{\mathbb{R}} \left\vert g(t) \right\vert {\text d}t < \infty , \) is convergent so that its Fourier transform exists. According to the Fourier integral theorem
\[ g(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-{\bf j} st} {\text d} s \left( \int_{-\infty}^{\infty} e^{{\bf j} sx} g(x) \,{\text d} x \right) . \]
Using definition of the function g(t), we find
\[ \frac{e^{ct}}{2\pi} \int_{-\infty}^{\infty} e^{-{\bf j} st} {\text d} s \left( \int_{-\infty}^{\infty} e^{{\bf j} sx} g(x) \,{\text d} x \right) = \begin{cases} f(t) , & \ \mbox{ for } t \ge 0, \\ 0, & \ \mbox{ for } t < 0. \end{cases} \]
So for t ≥ 0, we have
\[ f(t) = \frac{e^{ct}}{2\pi} \int_{-\infty}^{\infty} e^{-{\bf j} st} {\text d} s \left( \int_{-\infty}^{\infty} e^{{\bf j} sx - cx} f(x) \,{\text d} x \right) \]
If we set p = c − js, then
\[ f(t) = \frac{1}{2\pi{\bf j}} \,\int_{c-{\bf j}\infty}^{c+{\bf j}\infty} F(p)\,e^{pt} {\text d}p , \]
where
\[ F(p) = \int_0^{\infty} f(t)\,e^{-pt} {\text d}t . \]
Thus, in a purely formal manner, the transform formula
\[ F(p) = {\cal L}_{t\to p} \left[ f(t) \right] = \int_0^{\infty} f(t)\,e^{-pt} {\text d}t \]
is called the Laplace transform of f(t). and its inverse is
\[ f(t) = {\cal L}^{-1}_{p\to t} \left[ F(p) \right] = \frac{1}{2\pi{\bf j}} \,\int_{c-{\bf j}\infty}^{c+{\bf j}\infty} F(p)\,e^{pt} {\text d}p , \quad t> 0. \]
   ■
There known two other unitary versions of Fourier transforms:
\[ \hat{f}_1 (\xi ) = \int_{-\infty}^{\infty} f(t)\,e^{2\pi {\bf j} \xi\cdot t} \,{\text d}t , \qquad f (t) = \int_{-\infty}^{\infty} \hat{f}_1 (\xi ) \,e^{-2\pi {\bf j} \xi\cdot t} \,{\text d}\xi \]
and
\[ \hat{f}_2 (\xi ) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(t)\,e^{{\bf j} \xi\cdot t} \,{\text d}t , \qquad f (t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \hat{f}_2 (\xi ) \,e^{-{\bf j} \xi\cdot t} \,{\text d}\xi . \]
The latter is used in Mathematica under the name FourierTransform. There is no agreement on what particular definition of Fourier transform to be used. Of course, theoretical expositions of this topic prefer to use unitary versions.

Theorem 5: The Fourier transformation gives the spectral representation of the momentum operator \( \displaystyle \hat{p} = -{\bf j}\,\frac{\text d}{{\text d} x} , \) that is,
\[ ℱ \left[ {\bf j}\,f' (x) \right] = \int_{-\infty}^{\infty} {\bf j}\,\frac{{\text d}f}{{\text d}x}\,e^{{\bf j}x\xi} {\text d}x = \xi \int_{-\infty}^{\infty} f(x)\, e^{{\bf j}x\xi} {\text d}x = \xi \,ℱ \left[ f (x) \right] (\xi ). \]

This main property of the Fourier transform can be extended for multi-dimensional case.

Corollary: If a differentiable function is absolutely integrable, that is \( f \in 𝔏^1 \left( {\mathbb R}^n \right) , \) and \( \partial f/\partial x_j \in 𝔏^1 \left( {\mathbb R}^n \right) , \) then the Fourier transform of the derivative is
\[ ℱ \left[ {\bf j}\,\frac{\partial f}{\partial x_j} \right] (\xi ) = \xi_j ℱ \left[ f \right] (\xi ) = \xi_j \hat{f} (\xi ) , \qquad j=1,2,\ldots , n. \]

We will frequently use the following two norms:

\[ \| f \|_1 \equiv \| f \|_{𝔏^1} = \int | f (t) |\,{\text d}t \qquad \mbox{and} \qquad \| f \|_2 \equiv \| f \|_{𝔏^2} = \left( \int | f (t) |^2 \,{\text d}t \right)^{1/2} . \]
The integral \( \int_0^{\infty} f(x)\,{\text d}x \) is said to be Cesàro summable or (C.1) summable to the value I if
\[ \lim_{N\to \infty} \int_0^N \left( 1 - \frac{x}{N} \right) f(x) \,{\text d}x = I . \]

Theorem 6: Let f be absolutely integrable function, which we abbreviate as f ∈ 𝔏¹(ℝ). Then the following integral is (C.1) summable to
\[ \frac{1}{\pi} \int_0^{\infty} {\text d}u \int_{-\infty}^{\infty} f(t)\,\cos u \left( x - t \right) {\text d}t = \frac{f(x+0) + f(x-0)}{2} . \]

The cardinal sine function or unnormalized sinc function is defined by
\[ \mbox{sinc}(x) = \frac{\sin x}{x} = \prod_{n=1}^{\infty} \cos \left( \frac{x}{2^n} \right) . \]
In digital signal processing and information theory, the normalized sinc function is commonly defined as
\[ \mbox{sinc}_{\pi} (x) = \mbox{sinc} (\pi\, x) = \frac{\sin \pi x}{\pi\,x} = \prod_{n=1}^{\infty} \left( 1 - \frac{x^2}{n^2} \right) = \frac{1}{\Gamma (1+x)\,\Gamma (1-x)} , \]
where Γ is the Gamma function. In general, we have
\[ \mbox{sinc}_{\omega} (x) = \frac{\sin \omega x}{\omega\,x} = \mbox{sinc} \left( \omega \, x\right) . \]
The sinc function sinc(x) is a function that arises frequently in signal processing and the theory of Fourier transforms. Its inverse Fourier transform is called the "sampling function" or "filtering function." The full name of the function is "sine cardinal," but it is commonly referred to by its abbreviation, "sinc." In Mathematica, sinc function has a default notation: Sinc[x]. The zero crossings of the unnormalized sinc are at non-zero integer multiples of π, while zero crossings of the normalized sinc occur at non-zero integers.
sinc = Plot[Sinc[x], {x, -10, 10}, PlotStyle -> Thick, AxesLabel -> {x, Sinc[x]}];
arrow1 = Graphics[{Red, Arrowheads[0.05], Arrow[{{4.5, 0.5}, {3.14, 0.02}}]}];
arrow2 = Graphics[{Red, Arrowheads[0.05], Arrow[{{8.5, 0.7}, {9.4, 0.02}}]}];
arrow3 = Graphics[{Red, Arrowheads[0.05], Arrow[{{-4.5, 0.5}, {-6.28, 0.02}}]}];
text1 = Graphics[ Text[Style["\[Pi]", FontSize -> 14, Black], {4.4, 0.6}]];
text2 = Graphics[ Text[Style["3\[Pi]", FontSize -> 14, Black], {8.6, 0.8}]];
text3 = Graphics[ Text[Style["-2\[Pi]", FontSize -> 14, Black], {-4.4, 0.6}]];
Show[sinc, arrow1, arrow2, arrow3, text1, text2, text3]
Unnormalized sinc function Mathematica code

The antiderivative of sinc function defines a special function---sine integral---that is included into Mathematics as SinIntegral:

\[ \mbox{Si}(x) = \int_0^x \frac{\sin t}{t} \, {\text d}t = \int_0^x \mbox{sinc}(t) \, {\text d}t . \]
The sinc function is an even function, and its integral is
\[ \int_{-\infty}^{\infty} \mbox{sinc}(x)\,{\text d}x = \int_{-\infty}^{\infty} \frac{\sin x}{x}\,{\text d}x = 2 \int_{0}^{\infty} \mbox{sinc}(x)\,{\text d}x = \mbox{Si}(\infty ) = \pi . \]
Integrate[Sinc[x], {x, -Infinity, Infinity}]
\[Pi]
2*SinIntegral[Infinity]
\[Pi]
The function \( \displaystyle y(x) = a\,\mbox{sinc}(a\,x) = \frac{\sin ax}{x} \) is a bounded solution of the initial value problem for the following second order differential equation with regular singular point at the origin
\[ x\,\frac{{\text d}^2 y}{{\text d} x^2} + 2\, \frac{{\text d} y}{{\text d} x} + a^2 x\,y =0 \qquad y(0) = a, \quad y' (0) = 0. \]
For this function, we have
\[ \lim_{a\to \infty} \frac{\sin (a\,x)}{x} = \pi \,\delta (x) \qquad \Longleftrightarrow \qquad \lim_{a\to \infty} \int_{-\infty}^{\infty} f(x) \,\frac{\sin (a\,x)}{x} \,{\text d}x = \pi\, f(0) . \]
Mathematica confirms:
x*D[a*Sinc[a*x], {x, 2}] + 2*D[a*Sinc[a*x], {x}] + a*a*x*a*Sinc[a*x]
FullSimplify[%]
0
Limit[Integrate[(1 + x^2)^(-1) *a*Sinc[a*x], {x, -Infinity, Infinity}], a -> Infinity]
\[Pi]
The other solution \( \displaystyle \frac{\cos ax}{x} \) of the differential equation \( x\, y'' + 2\,y' + a^2 x\,y =0 \) is unbounded at x = 0, unlike its sinc function counterpart; obviously, sinc(0) = 1.

The sinc function is a Forier transform of the tent function

\[ f(x) = \begin{cases} 0, & \ \mbox{ for} \quad x < -w , \\ h \left( 1 - \frac{|x|}{w} \right) , & \ \mbox{ for} \quad |x| < w , \\ 0, & \ \mbox{ for} \quad x > w. \end{cases} \]
Using Mathematica, we find its Fourier transform
Simplify[Integrate[(1 + x/w)*Exp[I*x*t], {x, -w, 0}] + Integrate[(1 - x/w)*Exp[I*x*t], {x, 0, w}]]
-((E^(-I t w) (-1 + E^(I t w))^2)/(t^2 w))
\[ \hat{f}(\xi ) = - \frac{h}{w} \left( \frac{\sin w\xi}{\xi} \right)^2 . \]
Example 4: In applications, it is common to approximate functions with piecewise constant (or step) functions. Although these functions are simple, they are very important: they are used to approximate other more complicated functions. A piecewise function is a function that is defined by several subfunctions. If each piece is a constant function, then the piecewise function is called Piecewise constant function or Step function. Let us consider one of them:
\[ \Pi_{\omega} (t) = \begin{cases} 1 , & \ \mbox{for } \ -\omega < t < \omega , \\ 0 , & \ \mbox{otherwise}. \end{cases} \]
f0[x_] = Refine[Piecewise[{{1, -a < x < a}}], a > 0];
Plot[f0[x], {x, -4, 4}, Ticks -> None, AxesLabel -> {t, "\!\(\*SubscriptBox[\(f\), \(0\)]\)(t)"}, PlotStyle -> {Thickness[0.015]}, Epilog -> Style[{Text["-h/2", {-2, 0.1}], Text["h/2", {2, 0.1}]}, 14]]
Graph of step function
This step function is commonly referred to as filtering function because the multiplication by it leads to elimination of the high frequency contributions to the signal. Therefore, it is also called a low-pass filter. Its Fourier transform is
\[ \int_{-\omega}^{\omega} e^{{\bf j} x\cdot t} \,{\text d}t = 2\,\frac{\sin (\omega x)}{x} = 2\,\omega \,\mbox{sinc}_{\omega} (x) . \]
Using Mathematica, we get
Integrate[Exp[I*x*s], {x, -a, a}]
(2 Sin[a s])/s
Sinc[Pi/2]
2/\[Pi]
We verify the answers with standard Mathematica commands:
f0[x_] = Refine[Piecewise[{{1, -a < x < a}}], a > 0];
Sqrt[2*Pi]*Refine[FourierTransform[f0[x], x, s], a > 0]
(2 Sin[a s])/s
and
Refine[InverseFourierTransform[a*Sinc[a*x], x, t], a > 0]/Sqrt[2*Pi]
1/4 (Sign[a - t] + Sign[a + t])
Note that Mathematica uses the unitary definition for Fourier transformations and its inverse: \( \displaystyle f^F (\xi ) = \frac{1}{\sqrt{2\pi}} \int f(x)\,e^{{\bf j} x\cdot \xi} \,{\text d} x \) and \( \displaystyle f (x ) = \frac{1}{\sqrt{2\pi}} \int f^F (\xi )\,e^{-{\bf j} x\cdot \xi} \,{\text d} \xi . \)

Multiplying the Fourier transform \( \hat{f} (\xi ) \) by the low-pass filter Π(ξ) effectively clips all the high frequencies (those of frequencies that are > ξ). By taking the inverse transform of this product, we remove the contribution of the high frequencies of the signal f(x). Calculations show that

\[ f_n (x) = ℱ^{-1} \left[ \hat{f} (\xi )\,\Pi_n (\xi ) \right] = \frac{1}{2\pi} \,\int_{-\infty}^{\infty} \hat{f} (\xi )\,\Pi_n (\xi ) \, e^{-{\bf j}x\,\xi} \,{\text d} \xi . \]
According to the convolution theorem,
\[ f_n (x) = f\ast g_n (x) , \]
where
\[ g_n (x) = ℱ^{-1} \left[ \Pi_n (\xi ) \right] = \frac{1}{2\pi} \,\int_{-\infty}^{\infty} \Pi_n (\xi )\, e^{-{\bf j}x\,\xi} \,{\text d} \xi = \frac{1}{2\pi} \,\int_{-n}^n e^{-{\bf j}x\,\xi} \,{\text d} \xi = \frac{2}{2\pi} \,\int_{0}^n \cos (\xi x) \,{\text d} \xi = \frac{1}{\pi} \,\frac{\sin nx}{x} . \]
■

The Riemann-Lebesque Lemma: If f ∈ 𝔏¹(ℝ), then \( \displaystyle \lim_{\xi \to \infty} \,\left\vert \hat{f} (\xi ) \right\vert = 0 , \) so \( \displaystyle \lim_{\xi \to \infty} \,\int_{-\infty}^{\infty} f(t)\,e^{{\bf j} t\cdot \xi} \,{\text d} t = 0 . \)
The convolution theorem: If f, g : ℝ → ℝ are integrable and g is piecewise contintinuous. Then the convolution f ☆ g is integrable on ℝ and its Fourier transform is the product of their Fourier transforms:
\[ ℱ \left[ f\star g \right] = ℱ \left[ f \right] \cdot ℱ \left[ g \right] , \qquad\mbox{with} \quad \left( f\star g \right) (x) = \int f(x-t)\,g(t)\,{\text d}t . \]
Note that evaluation of the convolution integral (which is usually denoted either by ☆ or *) is an ill-posed problem.

Fourier transform of convolution is
\begin{align*} ℱ \left[ f\star g \right] &= \int_{-\infty}^{+\infty} \left( f \star g \right) (x)\, e^{{\bf j}x\xi} {\text d} x = \int_{-\infty}^{+\infty} {\text d} x\, e^{{\bf j}x\xi} \int_{-\infty}^{+\infty} f(s)\, g(x-s)\,{\text d}s \\ &= \int_{-\infty}^{+\infty} {\text d}s \, f(s)\, \int_{-\infty}^{+\infty} {\text d} x\, g(x-s)\,e^{{\bf j}x\xi} \\ &= \int_{-\infty}^{+\infty} {\text d}s \, f(s)\, \int_{-\infty}^{+\infty} {\text d} x\, g(x-s)\,e^{{\bf j}\left( x - s \right) \xi} e^{{\bf j}s\xi} \qquad (u \mapsto x-s) \\ &= \int_{-\infty}^{+\infty} {\text d}s \, f(s)\, e^{{\bf j}s\xi} \, \int_{-\infty}^{+\infty} {\text d} u\,g(u)\,e^{{\bf j}u\xi} \\ &= \hat{f} \cdot \hat{g} . \end{align*}

Example A: Let f be the exponential function \( \displaystyle f(x) = e^{-a|x|} , \) where a is a positive real number. Its Fourier stransform
\begin{align*} ℱ \left[ f\ast g \right] &= \int_{-\infty}^{\infty} e^{-a|x|} e^{{\bf j}x\xi} {\text d}x = \int_0^{\infty} e^{-a|x| + {\bf j}x\xi} {\text d}x + \int_0^{\infty} e^{-a|x| - {\bf j}x\xi} {\text d}x \\ &= \frac{1}{a-{\bf j}\xi} + \frac{1}{a+{\bf j}\xi} = \frac{a}{a^2 + \xi^2} , \quad a > 0. \end{align*}

 

Example B: We find the Fourier transform of the function \( \displaystyle f(x) = e^{-x^2 /2} : \)
\begin{align*} ℱ \left[ f \right] &= \int_{-\infty}^{\infty} e^{-x^2 /2} e^{{\bf j}x\xi} {\text d}x = e^{-\xi^2 /2} \int_{-\infty}^{\infty} e^{-(x-{\bf j}\xi )^2 /2} {\text d}x \\ &= 2\,e^{-\xi^2 /2} \int_{-\infty}^{\infty} e^{-u^2} {\text d}u = 2\sqrt{\pi} \, e^{-\xi^2 /2} \end{align*}
because
Integrate[Exp[-u^2], {u, -Infinity, Infinity}]
Sqrt[\[Pi]]
We can show it analytically:
\begin{align*} I^2 &= \left( \int_{-\infty}^{\infty} e^{-u^2}{\text d} u \right) \left( \int_{-\infty}^{\infty} e^{-v^2}{\text d} v \right) \\ &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-u^2 -v^2} {\text d}u\,{\text d}v \\ &= \int_0^{\infty}r\,{\text d}r\, e^{-r^2} \int_0^{2\pi} {\text d}\theta = \pi \end{align*}
when polar coordinates are used. Then
\[ I= \int_{-\infty}^{\infty} e^{-u^2}{\text d} u = \sqrt{\pi} . \]

 

Example C: Consider the piecewise function
\[ f(x) = \begin{cases} e^{-ax} , & \ x> 0 , \\ 0, & \ x< 0. \end{cases} \]
Its Fourier transform is
\begin{align*} ℱ \left[ f \right] &= \int_{0}^{\infty} e^{-a|x|} e^{{\bf j}x\xi} {\text d}x \\ &= \frac{1}{a-{\bf j}\xi} . \end{align*}

 

Example D: Let us find the Fourier transform of the Heaviside function
\[ ℱ \left[ H(t) \right] = \int_0^{\infty} e^{{\bf j}t\xi} {\text d}t = \lim_{B\to \infty} \int_0^{B} e^{{\bf j}t\xi} {\text d}t = \lim_{B\to \infty} \, \frac{e^{{\bf j}B\xi} - 1}{{\bf j}\xi} = - \frac{1}{{\bf j}\xi} + \lim_{B\to \infty} \, \frac{e^{{\bf j}B\xi}}{{\bf j}\xi} . \]
So the limit in the right-hand side does not exist because the exponential function \( e^{{\bf j}B\xi} \) has no limit at infinity. Now we find its limit in weak sense. Choosing a probe function ϕ, we multiply by it and integrate:
\[ \int_{-\infty}^{\infty} \frac{e^{{\bf j}B\xi} }{{\bf j}\xi} \,\phi (\xi )\,{\text d}\xi = \lim_{K\to \infty} \int_{-K}^{K} \frac{e^{{\bf j}B\xi} }{{\bf j}\xi} \,\phi (\xi )\,{\text d}\xi . \]
To find its value, we connect the end points -K and K on real axis ℝ with semi-circle in complex plane ℂ to make a close loop. Then we apply the residue theorem. Since we have half of a circle, we have only half of the residure:
\[ \int_{-\infty}^{\infty} \frac{e^{{\bf j}B\xi} }{{\bf j}\xi} \,\phi (\xi )\,{\text d}\xi = \pi{\bf j} \,\lim_{\xi \to 0} \frac{e^{{\bf j}B\xi} }{{\bf j}} \,\phi (\xi ) = \pi\, \phi (0) . \]
Therefore, we obtain the formula for Fourier transform of the Heaviside function:
\[ ℱ \left[ H(t) \right] (\xi ) = \pi\,\delta (\xi ) - \frac{1}{{\bf j}\xi} . \]
Similarly, we get
\[ ℱ \left[ H(-t) \right] (\xi ) = \pi\,\delta (\xi ) + \frac{1}{{\bf j}\xi} . \]
Since the signum function is the difference of two Heaviside functions
\[ \mbox{sign}(x) = H(x) - H(-x) = \begin{cases} 1, & \ \mbox{ for } x> 0, \\ 0 , & \ \mbox{ for } x=0, \\ -1 , & \ \mbox{ for } x < 0 , \end{cases} \]
we get its Fourier transform to be
\[ ℱ \left[ \mbox{sign}(x) \right] (\xi ) = \int_0^{\infty} e^{{\bf j}x\xi} {\text d}x - \int_{-\infty}^0 e^{{\bf j}x\xi} {\text d}x = \frac{2}{{\bf j}\xi} . \]

 

Example E: Let us find the Fourier transform of the following function
\[ f(x) = \begin{cases} 1- |x|, & \ \mbox{ for } |x|< 1, \\ 0 , & \ \mbox{ otherwise. } \end{cases} \]
Its Fourier transform is
\begin{align*} ℱ \left[ f(x) \right] &= \int_{-1}^1 \left( 1 - |x| \right) e^{{\bf j}\xi x} \,{\text d} x \\ &= 2 \int_0^1 \left( 1 - x \right) \cos (x\xi ) \,{\text d} x \\ &= 2 \left[ \frac{(1-x)\,\sin (x\xi )}{\xi} \right]_{x=0}^{x=1} + \frac{2}{\xi} \int_0^1 \sin (x\xi ) \,{\text d} x \\ &= \frac{4}{\xi^2} \,\sin^2 \left( \frac{\xi}{2} \right) . \end{align*}
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Example 6: Let f0(t) be the gate function:
\[ f_0 (t) = \begin{cases} 1, & \ -h/2 < t < h/2 , \\ 0, & \ \mbox{otherwise.} \end{cases} \]
Its Fourier transform was found in the previous example:
\[ ℱ \left[ f_0 \right] (\xi ) = \int_{-\h/2}^{h/2} e^{{\bf j} t\cdot \xi}\,{\text d}t = 2\, \frac{\sin \left( \frac{h\xi}{2} \right)}{\xi} = h\,\mbox{sinc}\left( \frac{h\xi}{2} \right) . \]
Let f1(t) be a convolution of two gate functions:
\[ f_1 (t) = \frac{1}{h}\, f_0 (t) \ast f_0 (t) = \frac{1}{h}\times \begin{cases} t+h , & \ -h < t \le 0 , \\ h-t , & \ 0 \le t < h , \\ 0, & \ \mbox{otherwise.} \end{cases} \]
Continuing this process, we define the sequence of functions given by
\[ f_{n-1} (t) = \frac{1}{h}\, f_0 (t) \ast f_{n-2} (t) , \qquad n= 2,3,\ldots . \]
Using convolution rule, it is not hard to verify that the Fourier transform of each function is the integer power of sinc function:
\[ ℱ \left[ f_{n-1} \right] (\xi ) = h\,\mbox{sinc}^n \left( \frac{h\xi}{2} \right) , \qquad n=1,2,3,\ldots . \]
Therefore, explicit expressions for functions in the defined above sequence of convolutions can be obtained by applying the inverse Fourier transform:
\[ f_{n-1} (t) = ℱ^{-1} \left[ h\,\mbox{sinc}^n \left( \frac{h\xi}{2} \right) \right] = \frac{2}{\pi} \int_0^{\infty} \frac{\sin^n x}{x^n}\,\cos \left( \frac{2xt}{h} \right) {\text d}x , \qquad n=1,2,3,\ldots . \]
However, the evaluation of fn-1(t) becomes progressively more difficult as the value of n increases. So we present two more terms in this sequence:
\begin{align*} f_2 (t) &= \frac{1}{2h^2} \times \begin{cases} \frac{9h^2}{4} + 3ht + t^2 , & \ -\frac{3h}{2} < t \le -\frac{h}{2} , \\ \frac{3 h^2}{2} - 2 t^2 , & \ -\frac{h}{2} < t \le \frac{h}{2} , \\ \frac{9h^2}{4} - 3ht + t^2 , & \ \frac{h}{2} < t \le \frac{3h}{2} , \\ 0, & \ \mbox{otherwise;} \end{cases} \\ f_3 (t) &= \frac{1}{6\,h^3} \times \begin{cases} 8h^3 + 12 h^2 t + 6ht^2 + t^3 , & \ -2h < t \le -h , \\ 4 h^3 -6ht^2 - 3 t^3 , & \ -h < t \le 0 , \\ 4 h^3 -6ht^2 + 3 t^3 , & \ 0 < t \le h , \\ 8h^3 - 12 h^2 t + 6ht^2 - t^3 , & \ h < t \le 2h , \\ 0, & \ \mbox{otherwise.} \end{cases} \end{align*}
We plot three first functions in this sequence.
f1[x_] = Refine[ Piecewise[{{1 + x/a, -a < x <= 0}, {1 - x/a, 0 < x < a}}], a > 0];
Plot[f1[x], {x, -4, 4}, Ticks -> None, AxesLabel -> {t, "\!\(\*SubscriptBox[\(f\), \(1\)]\)(t)"}, PlotStyle -> {Thickness[0.015]}, Epilog -> Style[{Text["-h", {-2, -0.05}], Text["h", {2, -0.05}]}, 14]]
f2[x_] = Piecewise[{{(9 + 6*x + x^2)/8, -3 < x <= -1}, {(6 - 2*x^2)/ 8, -1 < x < 1}, {(9 - 6*x + x^2)/8, 1 < x <= 3}}]
Plot[f2[x], {x, -5, 5}, Ticks -> None, AxesLabel -> {t, "\!\(\*SubscriptBox[\(f\), \(2\)]\)(t)"}, PlotStyle -> {Thickness[0.015]}, Epilog -> Style[{Text["-3h/2", {-3.2, 0.05}], Text["3h/2", {3.2, 0.05}]}, 14]]
f3[t_] = InverseFourierTransform[2*(Sinc[x])^3, x, t]/Sqrt[2*Pi]
1/16 (-(-3 + t)^2 Sign[-3 + t] + 3 (-1 + t)^2 Sign[-1 + t] - 3 (1 + t)^2 Sign[1 + t] + (3 + t)^2 Sign[3 + t])
Plot[f3[x], {x, -5, 5}, Ticks -> None, AxesLabel -> {t, "\!\(\*SubscriptBox[\(f\), \(3\)]\)(t)"}, PlotStyle -> {Thickness[0.015]}, Epilog -> Style[{Text["-2h", {-3.5, 0.08}], Text["2h", {3.5, 0.08}]}, 14]]
Function f1(t) Function f2(t) Function f3(t)
There are some properties of the functions fn(t):
  • fn(t) has degree n.
  • fn(t) is an even function of t.
  • fn(t) is a continuous function of t.
  • fn(t = 0 when \( |t| > n\,h/2, \) otherwise fn(t) ≥ 0.
  • The area under fn(t) is h.
■

 

Weak Convergence

Riemann--Lebesgue Lemma: If f ∈ 𝔏¹(ℝ), then
\[ \lim_{\nu \to \infty} \int_{-\infty}^{+\infty} f(x)\, e^{{\bf j} \nu x} {\text d} x = 0 . \]
Let us consider the forward Fourier transform
\[ \hat{f} (\nu ) = \int_{-\infty}^{+\infty} f(x)\, e^{{\bf j} \nu x} {\text d} x = - \int_{-\infty}^{+\infty} f(x)\, e^{{\bf j} \nu x + {\bf j}\pi} {\text d} x, \]
where we used the identity −1 = exp(jπ). Rewriting the integral as
\[ \hat{f} (\nu ) = - \int_{-\infty}^{+\infty} f(x)\, e^{{\bf j} \nu \left( x + \pi /\nu \right)} {\text d} x, \]
and changing variables by replacing x + π/ν by t, we obtain
\[ \hat{f} (\nu ) = - \int_{-\infty}^{+\infty} f\left( t - \frac{\pi}{\nu} \right) e^{{\bf j} \nu t} {\text d} x. \]
We may write combining this result with the initial representation of the Fourier transform, as
\[ \hat{f} (\nu ) = \frac{1}{2} \left[ \int_{-\infty}^{+\infty} f(t)\, e^{{\bf j}\nu t} {\text d} t - \int_{-\infty}^{+\infty} \left( t - \frac{\pi}{\nu} \right) e^{{\bf j}\nu t} {\text d} t \right] . \]
Applying the CBS inequality, we may write
\[ \left\vert \hat{f} (\nu ) \right\vert = \frac{1}{2} \left\vert \int_{-\infty}^{+\infty} f(t)\, e^{{\bf j}\nu t} {\text d} t - \int_{-\infty}^{+\infty} \left( t - \frac{\pi}{\nu} \right) e^{{\bf j}\nu t} {\text d} t \right\vert \le \frac{1}{2} \int_{-\infty}^{+\infty} \left\vert f(t) - f(t - \pi /\nu ) \right\vert {\text d} t \,\to \, 0 \]
as |ν| → ∞.
The Fourier transform of the Dirac delta-function is
\[ ℱ \left[ \delta (x) \right] (\xi ) = 1. \]
The Fourier transform of the Heaviside function H(t) is given by
\[ ℱ \left[ H (x) \right] (\xi ) = \frac{1}{{\bf j} \left( \xi - {\bf j}0 \right)} , \qquad \mbox{where} \quad {\bf j}^2 =-1. \]
Also, the Fourier transform of the principal value (1/x) is
\[ ℱ \left[ \mbox{P.V.} \frac{1}{x} \right] (\xi ) = -{\bf j}\pi\,\mbox{sign}(\xi ). \]
Moreover, taking the Fourier transform of the Heaviside function
\[ H(x) = \frac{1}{2} \left[ \mbox{sign}(x) +1 \right] , \]
we obtain the relation
\[ \frac{1}{x \pm {\bf j}0} =\mbox{P.V.} \left( \frac{1}{x} \right) \mp {\bf j}\pi \,\delta (x) , \]
known as the Lippmann--Schwinger equation.

 

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