ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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The scaled Gaussian integral and its parameter derivative

Example

For $a > 0$ , define $F (a) : = \int_{- \infty}^{\infty} e^{- a x^{2}} d x$ . Then

$F (a) = \sqrt{\frac{π}{a}}, F^{'} (a) = - \frac{\sqrt{π}}{2 a^{3 / 2}} = - \int_{- \infty}^{\infty} x^{2} e^{- a x^{2}} d x .$

Facts & Assumptions

Given: A positive parameter $a$ .

[L1]

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

[L2]

On an open domain and open parameter interval, if $f$ and $\partial_{t} f$ are continuous, one slice is absolutely improperly integrable, and $∣ \partial_{t} f ∣$ has an integrable bound uniform on each compact parameter interval, then $F^{'} (t) = \int_{D} \partial_{t} f (x, t) d x$ (Differentiation under an improper multiple integral under an integrable derivative bound).

[L3]

For real $α$ , $(x^{α})^{'} = α x^{α - 1}$ on $(0, \infty)$ (Continuity and derivatives of positive-base real powers).

[L4]

A monotone differentiable substitution preserves convergent improper integrals under the compact-truncation hypotheses (Change of variable in an improper integral).

[L5]

Exponential decay dominates every fixed polynomial power (The exponential dominates every fixed nonnegative integer power at $+ \infty$ ).

[L6]

The tail integral $\int_{1}^{\infty} x^{- 2} d x$ converges (The improper $p$ -test for rational exponents).

[L7]

A nonnegative function dominated on a tail by a function with convergent improper integral also has a convergent tail integral (Comparison tests for improper integrals).

Verification

technique · direct

1.1

The substitution $u = \sqrt{a} x$ is licensed by [L4], and [L1] gives $F (a) = a^{- 1 / 2} \int_{- \infty}^{\infty} e^{- u^{2}} d u = \sqrt{π} a^{- 1 / 2}$ .

L1L4algebra

1.2

Let $C \subset (0, \infty)$ be compact and put $m : = \min C > 0$ . Then $∣ \partial_{a} e^{- a x^{2}} ∣ = x^{2} e^{- a x^{2}} \leq x^{2} e^{- m x^{2}}$ for $a \in C$ . Applying [L5] with the variable $x^{2}$ shows this is eventually at most $x^{- 2}$ , so [L6] and [L7] make both tails integrable; continuity handles the compact middle interval. The slice at $a = 1$ is absolutely integrable by [L1]. Thus every hypothesis of [L2] holds on the open parameter interval $(0, \infty)$ and gives $F^{'} (a) = - \int_{- \infty}^{\infty} x^{2} e^{- a x^{2}} d x$ .

L1L2L5L6L7

2.1

Differentiating the explicit formula in step 1.1 with [L3] gives $F^{'} (a) = - (\sqrt{π} / 2) a^{- 3 / 2}$ , which combined with step 1.2 gives the displayed second-moment identity.

step 1.1step 1.2L3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

W. F. Trench, Functions Defined by Improper Integrals, Example 12 (standard reference, not scraped)
M. E. Taylor, Introduction to Analysis in Several Variables, §3.1 (standard reference, not scraped)