1 MATHEMATICA tutorial, Part 2.5: Fourier transform

Introduction to Linear Algebra with Mathematica

# Preface

This section is about a classical integral transformation, known as the Fourier transformation. 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 principle value regularization for its definition, as well as for its inverse. It gives the spectral decomposition of the derivative 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. It 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

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
$$\label{EqT.1} \hat{f} (\xi ) =ℱ\left[ f(x) \right] (\xi ) = f^F (\xi ) = \int_{-\infty}^{\infty} f(t)\,e^{{\bf j} \xi\cdot t} \,{\text d}t$$
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:
$$\label{EqT.3} \hat{f} (\xi ) = ℱ(f) = f^F = \int_{{\mathbb R}^n} f(t)\,e^{{\bf j} \xi\cdot t} \,{\text d}t, \quad f(t) = ℱ^{-1} \left( \hat{f} \right) = \frac{1}{(2\pi )^n} \int_{{\mathbb R}^n} \hat{f}(\xi )\,e^{-{\bf j} \xi\cdot t} \,{\text d}\xi , \qquad$$
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 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 sup norm defined by
$\| f \|_{\infty} = \sup_{x\in \mathbb{R}} \left\vert f(x) \right\vert .$
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) → 𝔏(ℝ) and $$\| f^{F} \|_{\infty} \le \| f \|_1 .$$
Observe that for any $$\xi \in {\mathbb R}^n ,$$
$\left\vert \hat{f} \right\vert \le \int | f (t) |\,{\text d}t .$
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Example 1: The Fourier transform of the lorentzian function with paraemter 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 residure at x = −j𝑎:
$f^F (\xi ) = -2\pi{\bf j}\,\mbox{Res}_{x = -{\bf j}a} \,\frac{a}{a^2 + x^2}\, e^{{\bf j}x\xi} = -2\pi{\bf j}\,\lim_{x\to -{\bf j}a} \,\frac{a}{2x} \, e^{{\bf j}x\xi} = \pi\,e^{-a|\xi |} .$
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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
$\left\langle f , g \right\rangle = \int_{-\infty}^{\infty} \overline{f}(t)\,g(t)\, {\text d}t ,$
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 principle value means that the limit is taken over symmetrical interval

$$\label{EqT.4} \text{P.V.} \int_{-\infty}^{\infty} \hat{f}(\xi )\,e^{-{\bf j} \xi\cdot t} \,{\text d}\xi = \lim_{N\to \infty} \, \int_{-N}^N \hat{f}(\xi )\,e^{-{\bf j} \xi\cdot t} \,{\text d}\xi$$
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 is 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
$$\label{EqT.5} \| \hat{f} \|_2^2 = \int_{-\infty}^{\infty} \left\vert \hat{f}(\xi ) \right\vert^2 {\text d}\xi = 2\pi\, \int_{-\infty}^{\infty} \left\vert f(x ) \right\vert^2 {\text d}x = 2\pi\, \| f \|^2 .$$

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:

$$\label{EqT.6} \| f \|_1 = \int_{-\infty}^{\infty} \left\vert f(x) \right\vert {\text d} x < \infty .$$
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:
$$\label{EqT.7} \| f \|^2_2 = \int_{-\infty}^{\infty} \left\vert f(x) \right\vert^2 {\text d} x < \infty .$$
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
• f is absolutely integrable.
• f is of bounded variation in any given compact interval.
• f must have a finite number of discontinuities in any given bounded interval, and the discontinuities cannot be infinite.
A sufficient condition for exitence 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
$$\label{EqT.8} \lim_{|\xi | \to \infty} \hat {f} (\xi ) = 0 .$$

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
$\frac{f(x+0) + f(x-0)}{2} = \frac{1}{\pi} \int_0^{\infty} {\text d}u \int_{-\infty}^{\infty} f(t)\,\cos u\left( x-t \right) {\text d}t .$
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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 2: 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 transfom 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 = cjs, 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
$f(t) = {\cal L}^{-1}_{p\to t} \left[ F(p) \right] = \frac{1}{2\pi{\bf j}} \,\int_{c-{\bf j}\infty}^{c+{\bf j}\infty} F(p)\,e^{pt} {\text d}p , \quad t> 0.$
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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 derivative operator $$\displaystyle {\bf j}\,\frac{\text d}{{\text d} x} ,$$ that is,
$ℱ \left[ {\bf j}\,f' (x) \right] = \int_{-\infty}^{\infty} {\bf j}\,\frac{{\text d}f}{{\text d}x}\,e^{{\bf j}x\xi} {\text d}x = \xi = \xi \int_{-\infty}^{\infty} f(x)\, e^{{\bf j}x\xi} {\text d}x = \xi \,ℱ \left[ f (x) \right] (\xi ).$
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 3: 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]]
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 main property of the Fourier transform is that it gives the spectral representation of the impulse operator $$\displaystyle T = {\bf j}\,\frac{\text d}{{\text d}x} .$$

Theorem 8: 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
$ℱ \left[ {\bf j}\,\frac{\partial f}{\partial x_j} \right] (\xi ) = \xi_j ℱ \left[ f \right] (\xi ) = \xi_j \hat{f} (\xi ) .$
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: The Fourier transform of the convolution of two functions is the product of their Fourier transforms:
$ℱ \left[ f\ast g \right] = ℱ \left[ f \right] \cdot ℱ \left[ g \right] , \qquad\mbox{with} \quad \left( f\ast g \right) (x) = \int f(x-t)\,g(t)\,{\text d}t .$
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*}
Example 5: 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.

Cosine and sine Fourier transforms

There are known two spectral representations for the product of the impulse operator $$\displaystyle \left( {\bf j} \,\frac{\text d}{{\text d}x} \right)^2 = - \frac{{\text d}^2}{{\text d}x^2} .$$ They can be derived from the main Fourier formula for either even function, f(-x) = f(x) or odd function, f(-x) = -f(x). Correspondingly, we obtain the cosine Fourier transformation and sine Fourier transformation.
For an integrable on the interval [0, ∞) function f, two transformations can be defined; one is called cosine Fourier transform:
$ℱ_c \left[ f \right] (s) = f^c (s) = \int_0^{\infty} f(x)\,\cos (sx) \,{\text d}x \qquad\mbox{and} \qquad f(x) = \frac{2}{\pi} \int_0^{\infty} ℱ_c \left[ f \right] (s)\,\cos (sx) \,{\text d}s ;$
and sine Fourier transform:
$ℱ_s \left[ f \right] (s) = f^s (s) = \int_0^{\infty} f(x)\,\sin (sx) \,{\text d}x \qquad\mbox{and} \qquad f(x) = \frac{2}{\pi} \int_0^{\infty} ℱ_s \left[ f \right] (s)\,\sin (sx) \,{\text d}s .$
These two transformations provide spectral representations for the second derivative operator with Dirichlet and Neumann boundary conditions, respectively. More precisely, we have
\begin{align*} ℱ_c \left[ \frac{{\text d}^2 f}{{\text d} x^2} \right] (k) &= \int_0^{\infty} f'' (x)\,\cos (kx) \,{\text d}x = - f' (0) - k^2 ℱ_c \left[ f \right] (k) , \\ ℱ_s \left[ \frac{{\text d}^2 f}{{\text d} x^2} \right] (k) &= \int_0^{\infty} f'' (x)\,\sin (kx) \,{\text d}x = k\, f (0) - k^2 ℱ_s \left[ f \right] (k) . \end{align*}
Mathematica has two dedicated commands to perform sine and cosine Fourier transforms: FourierSinTransform and FourierCosTransform; however, Mathematica defines its Fourier transforms as:
\begin{align*} {\bf{\text FourierSinTransform}}[f] &= \sqrt{\frac{2}{\pi}} \int_0^{\infty} f(x) \,\sin (kx) \,{\text d}x , \\ {\bf{\text FourierCosTransform}}[f] &= \sqrt{\frac{2}{\pi}} \int_0^{\infty} f(x) \,\cos (kx) \,{\text d}x . \end{align*}
Their inverse transforms become
\begin{align*} f(x) &= \sqrt{\frac{2}{\pi}} \int_0^{\infty}{\bf{\text FourierSinTransform}}[f] (k)\, \sin (kx)\,{\text d}k , \\ f(x) &= \sqrt{\frac{2}{\pi}} \int_0^{\infty}{\bf{\text FourierCosTransform}}[f] (k)\, \cos (kx)\,{\text d}k . \end{align*}
Upon introducing two generalized (not commutative) convolution rules
\begin{align*} \left( f \overset{0}{*} g \right) (x) &= \int_0^{\infty} f(y) \left[ g(|x-y|) + g(x+y) \right] {\text d}y , \\ \left( f \overset{1}{*} g \right) (x) &= \int_0^{\infty} f(y) \left[ g(|x-y|) - g(x+y) \right] {\text d}y , \end{align*}
we find their Fourier transforms:
\begin{align*} ℱ_c \left( f \overset{0}{*} g \right) (x) &= 2\,\left( ℱ_c \,f \right) \left( ℱ_c \,g \right) , \\ ℱ_s \left( f \overset{1}{*} g \right) (x) &= 2\,\left( ℱ_s \,f \right) \left( ℱ_s \,g \right) . \end{align*}
Example A: Consider the unit rectangular function
$r(x) = H(x-a) - H(x-b) = \begin{cases} 1 , & \ \mbox{ if } a < x < b , \\ 0, & \ \mbox{otherwise}, \end{cases} \qquad a < b,$
where H is the Heaviside step function. Its sine and cosine Fourier transforms are
\begin{align*} ℱ_s \left[ H (x-a) - H(x-b) \right] &= \int_a^{b} \sin (kx)\, {\text d}x = \frac{\cos (ak) - \cos (bk)}{k} , \\ ℱ_c \left[ H (x-a) - H(x-b) \right] &= \int_a^{b} \cos (kx)\, {\text d}x = \frac{\sin (bk) - \sin (ak)}{k} . \end{align*}
Integrate[Sin[k*x], {x, a, b}]
(Cos[a k] - Cos[b k])/k
Integrate[Cos[k*x], {x, a, b}]
(-Sin[a k] + Sin[b k])/k

Example B: The unit height tent function:
$f(x) = \begin{cases} x/a , & \ 0 < x < a, \\ (2a-x)/a, & \ a < x < 2a, \\ 0, & \ x > 2a. \end{cases}$
Its sine and cosine Fourier transforms are
\begin{align*} ℱ_s \left[ f(x) \right] &= \int_0^{2a} f(x)\,\sin (kx)\,{\text d}x = \frac{1}{a\,k^2}\left[ 2\,\sin (ak) - \sin (2ak) \right] , \\ ℱ_c \left[ f(x) \right] &= \int_0^{2a} f(x)\,\cos (kx)\,{\text d}x = \frac{4}{a\,k^2}\,\cos (ak)\,\sin^2 \frac{ak}{2} . \end{align*}
Integrate[x/a*Sin[k*x], {x, 0, a}] + Integrate[(2*a - x)/a*Sin[k*x], {x, a, 2*a}]
(2 Sin[a k] - Sin[2 a k])/(a k^2)
Integrate[x/a*Cos[k*x], {x, 0, a}] + Integrate[(2*a - x)/a*Cos[k*x], {x, a, 2*a}]
(4 Cos[a k] Sin[(a k)/2]^2)/(a k^2)

Example C: Consider an exponential function $$f(x) = e^{-\alpha\,x^2} .$$ Its sine and cosine Fourier transforms are
\begin{align*} ℱ_s \left[ e^{-\alpha\,x^2} \right] &= \int_0^{infty} e^{-\alpha\,x^2}\,\sin (kx)\,{\text d}x = \frac{1}{\sqrt{\alpha}}\,\mbox{DawnsonF} \left[ \frac{k}{2\sqrt{\alpha}} \right] ; \\ ℱ_c \left[ e^{-\alpha\,x^2} \right] &= \int_0^{\infty} e^{-\alpha\,x^2} \,\cos (kx)\,{\text d}x = \frac{\sqrt{\pi}}{2\sqrt{\alpha}}\, e^{-k^2 /(4\alpha )} , \end{align*}
where DawnsonF[x] is a special function
$\mbox{DawnsonF} [x] = e^{-x^2} \int_0^x e^{t^2}\,{\text d}t .$
Integrate[Exp[-a*x^2]*Sin[k*x], {x, 0, Infinity}, Assumptions -> a > 0 && k > 0]
DawsonF[k/(2 Sqrt[a])]/Sqrt[a]
Integrate[Exp[-a*x^2]*Cos[k*x], {x, 0, Infinity}, Assumptions -> a > 0 && k > 0]
(E^(-(k^2/(4 a))) Sqrt[Pi]])/(2 Sqrt[a]) Example D: We apply the Fourier transformations for sinc function: \begin{align*} ℱ_s \left[ \mbox{sinc}_a (x) \right] &= \int_0^{infty} \frac{\sin (ax)}{x} \,\sin (kx)\,{\text d}x = \frac{1}{2}\,\ln \left\vert \frac{a+k}{a-k} \right\vert ; \\ ℱ_c \left[ \mbox{sinc}_a (x) \right] &= \int_0^{\infty} \frac{\sin (ax)}{x} \,\cos (kx)\,{\text d}x = \begin{cases} \pi /2 , & \ \mbox{ if } k < a , \\ \pi /4 , & \ \mbox{ if } k = a , \\ 0 , & \ \mbox{ if } k > a. \end{cases} \end{align*} Integrate[Sin[a*x]/x*Sin[k*x], {x, 0, Infinity}, Assumptions -> a > 0 && k > 0] 1/2 Log[(a + k)/Abs[a - k]] Integrate[Sin[a*x]/x*Cos[k*x], {x, 0, Infinity}, Assumptions -> a > 0 && k > 0] 1/4 \[Pi] (1 + Sign[a - k]) Example A: For a positive number a, consider the Bessel function $$J_0 (ax)$$ of the first kind. Its cosine and sine Fourier transforms are \[ ℱ_c \left[ J_0 (ax) \right] = \int_0^{\infty} J_0 (ax)\,\cos (kx)\, {\text d} x = \begin{cases} \left( a^2 - k^2 \right)^{-1/2} , & \ \mbox{ if } 0 < k < a , \\ \infty , & \ \mbox{ if } k=a, \\ 0 , & \ \mbox{ if } k > a ; \end{cases}
$ℱ_s \left[ J_0 (ax) \right] = \int_0^{\infty} J_0 (ax)\,\sin (kx)\, {\text d} x = \begin{cases} 0, & \ \mbox{ if } 0 < k < a, \\ \left( k^2 - a^2 \right)^{-1/2} , & \ \mbox{ if } k > a. \end{cases}$
In general, we have
$ℱ_c \left[ J_{2n} (ax) \right] = \int_0^{\infty} J_{2n} (ax)\,\cos (kx)\,{\text d} x = \begin{cases} (-1)^n \left( a^2 - k^2 \right)^{-1/2} T_{2n} (k/a), & \ \mbox{ if } 0 < k < a , \\ \infty , & \ \mbox{ if } k=a, \\ 0 , & \ \mbox{ if } k > a ; \end{cases}$
$ℱ_s \left[ J_{2n+1} (ax) \right] = \int_0^{\infty} J_{2n+1} (ax)\, \sin (kx)\,{\text d} x = \begin{cases} (-1)^n \left( a^2 - k^2 \right)^{-1/2} T_{2n+1} (k/a), & \ \mbox{ if } 0 < k < a , \\ \infty , & \ \mbox{ if } k=a, \\ 0 , & \ \mbox{ if } k > a ; \end{cases}$
where T2n is the Chebyshev polynomial of the first kind. Another transforms:
$ℱ_c \left[ x^{-n} J_{2n} (ax) \right] = \int_0^{\infty} x^{-n} J_{2n} (ax)\,\cos (kx)\,{\text d} x = \begin{cases} \frac{\sqrt{\pi}}{\Gamma (n+1/2)} \left( 2a \right)^{-n} \left( a^2 - k^2 \right)^{n-1/2} , & \ \mbox{ if } 0 < k < a, \\ 0, & \ \mbox{ if } k > a , \end{cases}$
where Γ(x) is the gamma function.

Example B: Let $$\displaystyle He_n (x) = (-1)^n e^{x^2 /2} \frac{{\text d}^n}{{\text d} x^n} \left( e^{-x^2 /2} \right) , \quad n=0,1,2,\ldots ,$$ be the Hermite polynomial. The Fourier transforms are
$ℱ_c \left[ e^{-x^2 /2} He_{2n} (ax) \right] = \int_0^{\infty} e^{-x^2 /2} He_{2n} (ax)\,\cos (kx)\,{\text d} x = (-1)^n \sqrt{\frac{\pi}{2}}\, e^{-x^2 /2} k^{2n} ;$

Example C: Let $$\displaystyle L_n (x) = \frac{1}{n!}\, e^{x} \frac{{\text d}^n}{{\text d} x^n} \left( x^n e^{-x} \right) , \quad n=0,1,2,\ldots ,$$ be the Laguerre polynomial. The Fourier transforms are
$ℱ_c \left[ e^{-x^2 /2} L_{n} (x^2 ) \right] = \int_0^{\infty} e^{-x^2 /2} L_{n} (x^2 )\,\cos (kx)\,{\text d} x = \frac{1}{n!}\,\sqrt{\frac{\pi}{2}}\, e^{-k^2 /2} \left( He_n (k) \right)^2 .$
$ℱ_s \left[ e^{-x^2 /2} L_{n}^1 (x^2 ) \right] = \int_0^{\infty} e^{-x^2 /2} L_{n}^1 (x^2 )\,\sin (kx)\,{\text d} x = \frac{1}{n!}\,\sqrt{\frac{\pi}{2}}\, e^{-k^2 /2} \, He_n (k) \, He_{n+1} (k).$

For a real number μ, a generalized Fourier transform is defined as
$\hat{q} (k) = \int_0^{\infty} q(r)\left[ \cos (kr) - \frac{\mu}{k}\,\sin (kr) \right] {\text d}r .$
Its inverse transform is
$q (r) = \frac{2}{\pi} \int_0^{\infty} \hat{q} (k)\,\frac{k^2}{\mu^2 + k^2} \left[ \cos (kr) - \frac{\mu}{k}\,\sin (kr) \right] {\text d}k .$
Example 8: Consider the exponential function $$f(r) = e^{-\alpha\,r} .$$ Its generalized Fourier transform is
$\int_0^{\infty} \left[ \cos (kr) - \frac{\mu}{k}\,\sin (kr) \right] e^{-\alpha\,r} \, {\text d}r = \frac{\alpha - \mu}{\alpha^2 + k^2} , \qquad \Re\alpha > \Im k.$
Integrate[Exp[-alpha*r]*(Cos[k*r] - mu/k*Sin[k*r]), {r, 0, Infinity}]
ConditionalExpression[(alpha - mu)/(alpha^2 + k^2), Re[alpha] > Im[k]]
Integrate[ Exp[-alpha*r^2]*(Cos[k*r] - mu/k*Sin[k*r]), {r, 0, Infinity}]
ConditionalExpression[( E^(-(k^2/(4 alpha))) Sqrt[Pi]] (k - mu Erfi[k/(2 Sqrt[alpha])]))/( 2 Sqrt[alpha] k), Re[alpha] >= 0] Using previously determined formulas, we find \begin{align*} \int_0^{\infty} \left[ \cos (kr) - \frac{\mu}{k}\,\sin (kr) \right] e^{-\alpha\,r^2} \, {\text d}r &= \int_0^{\infty} \cos (kr) \, e^{-\alpha\,r^2} \, {\text d}r - \frac{\mu}{k} \,\int_0^{\infty} \sin (kr) \, e^{-\alpha\,r^2} \, {\text d}r \\ &= \end{align*} The Hankel transform of order ν of a function f(r) is given by \[ F_{\nu} (k) = \int_0^{\infty} f(r)\,J_{\nu} (kr)\,r\,{\text d}r ,
where Jν is the Bessel function of the first kind of order ν with ν ≥ −1/2. The inverse Hankel transform of Fν(k) is defined as
$f(r) = \int_0^{\infty} F_{\nu} (k)\,J_{\nu} (kr)\,k\,{\text d}k .$
For Hankel transformations, we have
$\int_0^{\infty} r\left( \frac{{\text d}^2 f}{{\text d} r^2} + \frac{1}{r} \, \frac{{\text d} f}{{\text d} r} - \frac{\nu^2}{r^2} \, f \right) J_{\nu} (rk)\,{\text d}r = - k^2 F_{\nu} (k) = - k^2 \int_0^{\infty} f(r)\,J_{\nu} (kr)\,r\,{\text d}r .$

Weak Convergence

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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