---
title: "The Fourier Transform on the Line"
description: "Extend Fourier series from an interval to the whole line, where the sum over discrete frequencies becomes an integral over a continuum."
canonical: https://lightmysky.com/learn/mathematics/the-fourier-transform-on-the-line-mt_BMrt5FPe3U
source: https://lightmysky.com/learn/mathematics/the-fourier-transform-on-the-line-mt_BMrt5FPe3U.md
retrieved: 2026-11-12
---

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# 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.

Subject: Mathematics · Area: Calculus & Analysis · Ages 22 to 23
Page: https://lightmysky.com/learn/mathematics/the-fourier-transform-on-the-line-mt_BMrt5FPe3U

## Ready when they can

- 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

## Lesson: From repeating tones to a continuum

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.

**Example.** 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.

**Example.** 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.

**Example.** 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.

*(drawing: Under the jump-point theorem, limits 3 and 9 give an inversion limit of 6.)*

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.

**Recap.** 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.*

## Practice

14 questions on this page, each with its working shown.

## Needs first

- [Fourier Series](https://lightmysky.com/learn/mathematics/fourier-series-mt_DJ3iwfj7NK)
- [Lp Spaces and the Inequalities They Rest On](https://lightmysky.com/learn/mathematics/lp-spaces-and-the-inequalities-they-rest-on-mt_Qd8jU26qgE)

## Opens up

- [Convolution, Plancherel and Transforming a Derivative](https://lightmysky.com/learn/mathematics/convolution-plancherel-and-transforming-a-derivative-mt_zSwMJrQIMt)
