The Fourier Transform on the Line
Extend Fourier series from an interval to the whole line, where the sum over discrete frequencies becomes an integral over a continuum.
What a learner can do afterwards
- Derive the Fourier integral from periodized Fourier series on intervals of growing length for smooth, rapidly decreasing functions
- Compute the transform of a Gaussian and of an indicator function
- State the inversion theorem and the hypotheses it needs
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Use frequency xi in cycles per unit. Define fhat(xi) = ∫_R f(x) exp(-2pi i xi x) dx. Here R means the whole real line, i^2 = -1, and exp(z) means e^z. If f is in L1, meaning ∫_R |f(x)| dx is finite, the transform integral converges absolutely at every xi. The inverse transform uses the opposite exponential sign.
Start with a Schwartz function: f is smooth, and f and every derivative decay faster than every inverse power of |x| at infinity. Its periodization is P_L(x) = sum over integers k of f(x+kL). For integer n, the complex Fourier coefficient is c_n = (1/L)∫ from -L/2 to L/2 of P_L(x) exp(-2pi i n x/L) dx. Unfolding the translated intervals gives c_n = fhat(n/L)/L.
The series for P_L is sum over integers n of fhat(n/L) exp(2pi i (n/L)x)/L. Its frequencies xi_n = n/L have spacing Delta xi = 1/L. Each term includes this spacing, making a Riemann sum. For Schwartz f, as L grows, P_L(x) tends to f(x) and the sum tends to ∫_R fhat(xi) exp(2pi i xi x) dxi. Doubling L halves the spacing.
Let g(x) = exp(-pi a x^2), a > 0, and G = F(g). The Gaussian integral gives G(0) = 1/sqrt(a). Since g' = -2pi a xg, differentiation under the integral gives G' = (i/a)F(g'). Integration by parts gives F(g') = 2pi i xi G. Thus G' = -(2pi xi/a)G. Solving with G(0) gives G(xi) = a^(-1/2) exp(-pi xi^2/a). Gaussian decay justifies these steps.
For a = 2, exp(-2pi x^2) transforms to (1/sqrt(2)) exp(-pi xi^2/2). Increasing a narrows the input Gaussian and widens its transform. At a = 1, exp(-pi x^2) is its own transform under this convention. Other values of a change its scale.
For the indicator of [-a,a], a > 0, integrate exp(-2pi i xi x) from -a to a. For xi != 0, an antiderivative is exp(-2pi i xi x)/(-2pi i xi). Subtracting endpoint values gives fhat(xi) = sin(2pi a xi)/(pi xi). At xi = 0, the defining integral gives fhat(0) = 2a, also the continuous value of the sinc expression. Changing finitely many endpoint values leaves every integral unchanged.
If G is the transform of g, substituting u = x-c shows that g(x-c) transforms to exp(-2pi i c xi)G(xi). For the indicator of [-1,5], the center is c = 2 and half-width is a = 3. Its transform is exp(-4pi i xi) sin(6pi xi)/(pi xi) for xi != 0, with value 6 at zero. Translation adds a frequency-dependent phase.
One inversion theorem assumes both f and fhat are in L1. At every continuity point x of f, it gives f(x) = ∫_R fhat(xi) exp(2pi i xi x) dxi. This inverse integral converges absolutely. The positive exponential sign undoes the negative sign in the forward transform. These hypotheses are sufficient for this theorem; other inversion theorems use different assumptions.
Bounded variation means sums of absolute changes over ordered partitions have a common finite bound. If f is in L1 and has bounded variation near x, then lim_(R->infinity) ∫ from -R to R of fhat(xi) exp(2pi i xi x) dxi = (f(x-)+f(x+))/2. Here f(x-) and f(x+) are the one-sided limits. At continuity points this is f(x); at jumps it is their midpoint, whatever value is assigned at x.
For Schwartz functions, two forward transforms give F(Ff)(x) = f(-x). Inversion uses the opposite exponential sign instead. Even functions hide this reflection. For example, exp(-pi (x-2)^2) is reflected to exp(-pi (x+2)^2), moving its center from 2 to -2.
Suppose f is continuously differentiable, f and f' are in L1, and f tends to zero at both infinities. Integration by parts removes the boundary term and gives F(f')(xi) = 2pi i xi fhat(xi). Differentiation has become multiplication by 2pi i xi. This can turn a differential equation into an algebraic equation in frequency, followed by inversion when its hypotheses hold.
The ordinary Fourier integrals here do not work for every function without hypotheses. For f(x) = 1 on the whole line, the forward integral at xi = 0 is ∫_R 1 dx, which diverges. The failure occurs before inversion. Generalized transforms use another framework.
Series coefficients include frequency spacing. Compute transforms with one convention, check inversion hypotheses, and distinguish the inverse from two forward transforms.
Original lesson, questions and flashcards written for this dataset.
Original readings, questions and flashcards written for this dataset. Linked videos: 'Fourier Series introduction' and 'First term in a Fourier series'. Video licenses are not specified.
More to read
The frequency-grid weight
Fix fhat(xi) = ∫_R f(x) exp(-2pi i xi x) dx, with xi measured in cycles per unit. For f in L1, the kernel has modulus one, so this integral converges absolutely.
Take a Schwartz function: f is smooth, and f and all its derivatives decay faster than every inverse power of |x| at infinity. Adding its translates f(x+kL) over integers k gives a period-L function P_L.
The nth complex Fourier coefficient of P_L is fhat(n/L)/L. To see this, insert the sum of translates into the coefficient integral over one period. In each term substitute u = x+kL. The extra factor is exp(2pi i nk), which equals one. The translated intervals cover the real line, leaving the whole-line transform integral divided by L.
The Fourier series samples frequencies n/L and weights each term by 1/L. This is the frequency-grid width. For Schwartz functions, letting L grow makes the weighted sum approach the inverse Fourier integral, while P_L approaches f. The shrinking gaps alone would not justify that limit for arbitrary functions.
Computing the Gaussian transform
Let g(x) = exp(-pi a x^2), where a > 0, and write its transform as G. First find G(0), the integral of g. For a = 1, square this positive integral and use polar coordinates. Its square becomes 2pi∫ from 0 to infinity of r exp(-pi r^2) dr = 1. Substituting u = sqrt(a)x then gives G(0) = 1/sqrt(a).
Gaussian decay permits differentiation under the transform integral. Since g'(x) = -2pi a xg(x), differentiation with respect to xi gives G'(xi) = (i/a)F(g')(xi). Integration by parts has a zero boundary term and gives F(g')(xi) = 2pi i xi G(xi). Combining these equations yields G'(xi) = -(2pi xi/a)G(xi).
Solving this differential equation with its value at zero gives G(xi) = a^(-1/2) exp(-pi xi^2/a). This derivation stays on the real line. The case a = 1 is unchanged by the transform; other values change its width and amplitude.
The two linked enrichment videos review periodic Fourier series. They are prerequisite refreshers, not demonstrations of this Gaussian computation or whole-line inversion.
Intervals, zero frequency and translation
The indicator of [-a,a] equals one on that interval and zero outside it. For a > 0, its Fourier transform is therefore the integral of exp(-2pi i xi x) from -a to a.
When xi is nonzero, an antiderivative is exp(-2pi i xi x)/(-2pi i xi). Subtract the endpoint values and use the difference-of-exponentials identity. The result is sin(2pi a xi)/(pi xi).
At zero frequency, return to the defining integral. It equals the interval length, 2a. The sinc expression tends to 2a as xi tends to zero, so its apparent singularity is removable. Including or excluding either endpoint makes no difference to the integral because those points form a null set.
For an interval centered at c, substitute u = x-c. The kernel separates into exp(-2pi i c xi) times exp(-2pi i xi u). The transform is exp(-2pi i c xi) sin(2pi a xi)/(pi xi) away from zero, with value 2a at zero. The half-width a controls the sinc factor; the center c controls the phase. Translation preserves the transform's magnitude.
Inversion and reflection
The forward kernel is exp(-2pi i xi x), and the inverse kernel has the positive sign. If f and fhat are both in L1, the inverse integral converges absolutely and recovers f(x) at every continuity point. Smoothness is not an extra requirement of this theorem.
A second theorem uses symmetric truncation. If f is in L1 and has bounded variation near x, integrate fhat(xi) exp(2pi i xi x) over [-R,R], then let R tend to infinity. The limit is (f(x-)+f(x+))/2. It gives the function value at continuity points and the midpoint of the one-sided limits at jumps. Changing a single assigned point value does not change the transform.
These are different sufficient conditions for different inversion statements. Neither establishes ordinary-integral inversion for every function.
For Schwartz functions, two forward transforms give f(-x). Even functions conceal the reflection, but shifting a Gaussian away from zero makes it visible. Applying the inverse after the forward transform returns the original instead.
The constant function one shows why the framework matters: its ordinary transform integral diverges at zero frequency. Generalized transforms require additional definitions.
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