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Tonelli and the plane polar formula give int_R e^{-x^2} dx = sqrt(pi)

Example

Let $I : = \int_{R} e^{- x^{2}} d x .$ Then $I = \sqrt{π}$ .

Facts & Assumptions

Given: The nonnegative function $x \mapsto e^{- x^{2}}$ on $R$ .

[L1]

Tonelli's theorem applies to nonnegative functions on $R^{2}$ . (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product)

[L2]

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

Verification

technique · direct

1.1

Since $e^{- x^{2}} e^{- y^{2}} = e^{- (x^{2} + y^{2})}$ is nonnegative, Tonelli gives $I^{2} = \int_{R} \int_{R} e^{- (x^{2} + y^{2})} d y d x .$

2.1

By [L2], the double integral in step 1.1 is $π$ . Since $I \geq 0$ , it follows that $I^{2} = π$ and hence $I = \sqrt{π}$ .

L2step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

31 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

Gerald B. Folland, Real Analysis, 2nd ed., Proposition 2.53 (standard reference, not scraped)