Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-21
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The square of the one-dimensional Gaussian integral is the plane Gaussian integral

Statement

Let I=∫−∞∞e−x2 dx. Then

I2=∫R2e−(x2+y2) d(x,y).

The plane integral is the nonnegative improper multiple integral.

Facts & Assumptions

Given: The finite positive number I and the nonnegative plane Gaussian.

[L1]

For continuous a and b on rectangles, ∫A×Ba(x)b(y)=(∫Aa)(∫Bb) (The integral of a product function on a product rectangle is the product of the two integrals).

[L2]

Every compact Jordan exhaustion computes the nonnegative improper integral, independently of the exhaustion (Every Jordan exhaustion computes a nonnegative improper multiple integral).

[L3]

The integral I=∫−∞∞e−x2 dx exists as a finite positive real number (The improper integral of e−x2 over R is finite and positive).

[L4]

The exponential satisfies exp⁡(u+v)=exp⁡(u)exp⁡(v) (The exponential addition formula exp⁡(x+y)=exp⁡(x)exp⁡(y)).

Proof

technique · direct
1.1L1L4

For R>0, [L4] gives e−(x2+y2)=e−x2e−y2, so [L1] yields ∫[−R,R]2e−(x2+y2) d(x,y)=(∫−RRe−x2 dx)2.

2.1step 1.1L2L3

Put R=j+1. The squares form a compact Jordan exhaustion of R2, so [L2] makes their plane integrals tend to the nonnegative improper plane integral; [L3] makes each one-dimensional factor tend to I.

3.1step 2.1algebra∎

Passing to the limit in the product identity of step 1.1 gives the plane integral equal to I2.

Depends on

Used by

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Sources