TheoremStatement: Literature-sourcedProof: 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 Gaussian integral $\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π}$

Statement

$\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π} .$

Facts & Assumptions

Given: Write $I : = \int_{- \infty}^{\infty} e^{- x^{2}} d x$ .

[L1]

The integral $I$ exists as a finite positive real number (The improper integral of $e^{- x^{2}}$ over $R$ is finite and positive).

[L2]

One has $I^{2} = \int_{R^{2}} e^{- (x^{2} + y^{2})} d (x, y)$ (The square of the one-dimensional Gaussian integral is the plane Gaussian integral).

[L3]

The plane integral equals $π$ (The plane Gaussian integral equals $π$ by polar coordinates).

[L4]

Every nonnegative real has a unique nonnegative square root (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

Proof

technique · direct

1.1

By [L1], $I > 0$ , and [L2] with [L3] gives $I^{2} = π$ .

L1L2L3

2.1

Since $I$ is nonnegative, uniqueness in [L4] identifies it with $\sqrt{π}$ .

step 1.1L4∎

Depends on

Used by

The scaled Gaussian integral and its parameter derivative Example

Dependency tree · two levels

43 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

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

The Gaussian integral ∫−∞∞e−x2 dx=π∫−∞∞​e−x2dx=π​