LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck pass

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

First and second Gaussian heat-kernel moments

Statement

Assume Countable Choice. For $n \geq 1$ and $t > 0$ , all first and second moments are absolutely integrable and $\int_{R^{n}} x_{i} Γ (x, t) d x = 0, \int_{R^{n}} x_{i} x_{j} Γ (x, t) d x = 2 t δ_{i j} .$ In particular $\int_{R^{n}} ∣ x ∣^{2} Γ (x, t) d x = 2 n t$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $t > 0$ , and coordinate indices $0 \leq i, j < n$ wherever they appear.

[A1]

Countable Choice is the hypothesis carried by the integration and change-of-variables suppliers below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

For $t > 0$ the heat kernel is $Γ (x, t) = (4 π t)^{- n / 2} \exp (- ∣ x ∣^{2} / (4 t)) > 0$ on $R^{n}$ (The heat kernel on $R^{n}$ and its causal extension).

[F2]

$\int_{- \infty}^{\infty} e^{- u^{2}} d u = \sqrt{π}$ (The Gaussian integral $\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π}$ ).

[F3]

Under $R^{m + n} = R^{m} \times R^{n}$ the Lebesgue measure $λ_{m + n}$ is the completion of the product measure $λ_{m} \times λ_{n}$ (The Euclidean Lebesgue measure is the completion of the product of the factor Lebesgue measures).

[F4]

On completed sigma-finite product measure spaces, Tonelli applies to nonnegative completed-product-measurable functions and Fubini to $L^{1}$ functions. Sections are measurable (and in the Fubini case integrable) outside measurable null sets; define their inner integrals to be zero on those exceptional sets before taking the outer integral (Tonelli and Fubini for the completed product, with only almost-everywhere section measurability). The Gaussian products and moment integrands used here have measurable sections everywhere; their one-dimensional factors are integrable, so their ordinary iterated integrals agree with these modified integrals.

[F5]

For a $C^{1}$ diffeomorphism $T : U \to V$ of open sets and every $f \in L^{1} (V)$ , $\int_{V} f (y) d y = \int_{U} f (T (x)) ∣ \det D T (x) ∣ d x$ ; in particular for $n = 1$ both $u \mapsto - u$ and $u \mapsto c u$ with $c > 0$ qualify (A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions, A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not).

[F6]

For every $m \in N$ and real $a > 0$ , $s^{m} / \exp (a s) \to 0$ as $s \to + \infty$ (The exponential dominates every fixed nonnegative integer power at $+ \infty$ ).

[F7]

If $F, G$ are continuous on $[a, b]$ and differentiable on $(a, b)$ with $F^{'} = f$ , $G^{'} = g$ Riemann integrable there, then $\int_{a}^{b} F g + \int_{a}^{b} f G = F (b) G (b) - F (a) G (a)$ (Integration by parts for continuous factors with Riemann-integrable extensions of their interior derivatives).

[F8]

If $T$ preserves a measure $μ$ and $f$ is integrable, then $\int f \circ T d μ = \int f d μ$ (Integral invariance under measure-preserving maps); the reflection $s \mapsto - s$ preserves Lebesgue measure on $R$ .

Proof

technique · direct

1.1A1F1F2F3F4F5F6givenalgebra

Work under [A1] and fix $t > 0$ . For $m \in {1, 2}$ set $K_{m} (t) : = \sup_{s \in R} ∣ s ∣^{m} e^{- s^{2} / (8 t)} < \infty$ : the function is continuous and tends to $0$ at infinity by [F6] applied to the radial variable, so the supremum is finite; consequently, by [F1], $∣ x_{i} ∣^{m} Γ (x, t) \leq (4 π t)^{- n / 2} K_{m} (t) \prod_{k < n} e^{- x_{k}^{2} / (8 t)}$ . By the identification [F3] and Tonelli's theorem [F4], $\int_{R^{n}} \prod_{k} e^{- x_{k}^{2} / (8 t)} d x = \prod_{k} \int_{R} e^{- x_{k}^{2} / (8 t)} d x_{k} = (\sqrt{8 π t})^{n} < \infty$ , where each one-dimensional factor is computed by the substitution $x_{k} = 2 \sqrt{2 t} u$ of [F5] and the Gaussian integral [F2]; hence $∣ x_{i} ∣^{m} Γ (\cdot, t) \in L^{1} (R^{n})$ for $m = 1, 2$ and every $i$ , and $∣ x_{i} x_{j} ∣ \leq (x_{i}^{2} + x_{j}^{2}) / 2$ gives absolute integrability of the mixed moments as well, so Fubini's clause of [F4] applies to them.

2.1F4F5F8step 1.1givenalgebra

First moments: by Fubini's theorem [F4] applied to the integrable function $x \mapsto x_{i} Γ (x, t)$ , the integral is the iterated integral in which the $i$ -th factor is $\int_{R} s e^{- s^{2} / (4 t)} d s$ ; the function $s \mapsto s e^{- s^{2} / (4 t)}$ is odd and integrable, so by the change of variables $s = - u$ of [F5] its integral equals its own negative and is therefore $0$ , while all other factors are finite by step 1.1; hence $\int x_{i} Γ (x, t) d x = 0$ .

2.2F4F5F8step 1.1givenalgebra

Off-diagonal second moments: for $i \neq j$ , Fubini [F4] applied to the integrable $x \mapsto x_{i} x_{j} Γ (x, t)$ factors the integral into the product of the one-dimensional integrals $\int_{R} s e^{- s^{2} / (4 t)} d s$ in the $i$ -th and $j$ -th coordinates and the finite Gaussian factors in the remaining coordinates; each of the two odd factors vanishes by the change of variables $s = - u$ of [F5], so $\int x_{i} x_{j} Γ (x, t) d x = 0$ for $i \neq j$ .

2.3step 1.1F2F4F5F6F7givenalgebra

Diagonal second moments: fix $i$ and put $F (s) = s$ , $G (s) = e^{- s^{2} / (4 t)}$ on $[- R, R]$ ; integration by parts [F7] gives $\int_{- R}^{R} s^{2} e^{- s^{2} / (4 t)} d s = 2 t \int_{- R}^{R} e^{- s^{2} / (4 t)} d s - 4 t R e^{- R^{2} / (4 t)}$ . Letting $R \to \infty$ , the boundary term tends to $0$ by [F6] and the remaining integral equals $\sqrt{4 π t}$ by the substitution $s = 2 \sqrt{t} u$ of [F5] and [F2], so $\int_{R} s^{2} e^{- s^{2} / (4 t)} d s = 2 t \sqrt{4 π t}$ ; Fubini [F4] applied to the integrable function $x \mapsto x_{i}^{2} Γ (x, t)$ now gives $\int_{R^{n}} x_{i}^{2} Γ (x, t) d x = (4 π t)^{- n / 2} \cdot 2 t \sqrt{4 π t} \cdot (\sqrt{4 π t})^{n - 1} = 2 t$ .

3.1step 1.1step 2.1step 2.2step 2.3algebra∎

Steps 1.1, 2.1, 2.2 and 2.3 give absolute integrability, vanishing first moments, the covariance identity $\int x_{i} x_{j} Γ = 2 t δ_{i j}$ for every pair $i, j$ , and, summing the $n$ diagonal identities by linearity of the integral, $\int ∣ x ∣^{2} Γ (x, t) d x = \sum_{i < n} 2 t = 2 n t$ .

Depends on

Used by

Dependency tree · two levels

96 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

John K. Hunter, Notes on Partial Differential Equations (revised 18 June 2014, UC Davis) (standard reference, not scraped)
Jared Speck, MIT 18.152 Introduction to Partial Differential Equations, Class Meeting #5: The Fundamental Solution for the Heat Equation (Fall 2011) (standard reference, not scraped)