TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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Exact polynomial moments for the linear Fourier phase

Statement

For real endpoints $a < b$ , $ω \in R ∖ {0}$ , and an integer $m \geq 0$ , the output of Exact recurrence for polynomial Fourier moments satisfies $μ_{k} = \int_{a}^{b} x^{k} e^{i ω x} d x$ for every $0 \leq k \leq m$ , with $μ_{0} = \frac{e^{i ω b} - e^{i ω a}}{i ω}, μ_{k} = \frac{b^{k} e^{i ω b} - a^{k} e^{i ω a}}{i ω} - \frac{k}{i ω} μ_{k - 1} (k \geq 1) .$ Thus all polynomial moments for the Fourier phase $g (x) = x$ are obtained by a finite exact recurrence.

Facts & Assumptions

Given: Reals $a < b$ , $ω \neq 0$ , and the componentwise complex integral convention of Finite-interval oscillatory integral and phase convention.

[F1]

Exact recurrence for polynomial Fourier moments: $μ_{0} = (e^{i ω b} - e^{i ω a}) / (i ω)$ and $μ_{k} = (b^{k} e^{i ω b} - a^{k} e^{i ω a}) / (i ω) - k μ_{k - 1} / (i ω)$ .

[F2]

Derivative of a complex oscillatory phase: $(e^{i ω x})^{'} = i ω e^{i ω x}$ .

[F3]

If $u, v$ are differentiable on $[a, b]$ with $u^{'}, v^{'}$ integrable, then $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ : $\int u v^{'} = [u v] - \int u^{'} v$ for real $C^{1}$ functions; the componentwise complex identity follows by applying it to real and imaginary parts.

[F4]

For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term: for $k \geq 1$ , $(x^{k})^{'} = k x^{k - 1}$ .

Proof

1.1givenF1F2F3algebra

The algorithm sets $μ_{0} = (e^{i ω b} - e^{i ω a}) / (i ω)$ . Since $(e^{i ω x} / (i ω))^{'} = e^{i ω x}$ by [F2], [F3] with $u = 1$ identifies this value with $\int_{a}^{b} e^{i ω x} d x$ .

2.1givenF1F2F3F4step 1.1algebra∎

For $k \geq 1$ , apply [F3] to $u = x^{k}$ and $v^{'} = e^{i ω x}$ . Using [F2] and [F4] gives $\int_{a}^{b} x^{k} e^{i ω x} d x = (b^{k} e^{i ω b} - a^{k} e^{i ω a}) / (i ω) - k \int_{a}^{b} x^{k - 1} e^{i ω x} d x / (i ω)$ . Starting from step 1.1, induction identifies each value returned by the algorithm with the corresponding moment.

Depends on

Used by

A Fourier integral missed by one sample but captured by Filon moments Example

Dependency tree · two levels

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Sources

A. Iserles, S. P. Nørsett, and S. Olver, Highly oscillatory quadrature: The story so far, §3.2 (standard reference, not scraped)
D. Levin, Procedures for computing one- and two-dimensional integrals of functions with rapid irregular oscillations, §1 (standard reference, not scraped)
NIST Digital Library of Mathematical Functions, §2.3(i) (standard reference, not scraped)