ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck 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

1.1L1L4algebra

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

1.2L1L2L5L6L7

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 $F^{'} (a) = - \int_{- \infty}^{\infty} x^{2} e^{- a x^{2}} d x$ .

2.1step 1.1step 1.2L3∎

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.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

48 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

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)