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Γ(1/2)=π from the Gaussian integral

Statement

Γ(1/2)=π.

Facts & Assumptions

Given: Euler's Gamma integral at s=1/2.

[F1]

If ϕ:IJ is a monotone differentiable surjection with locally integrable derivative, the proper change-of-variable hypotheses hold on every compact truncation, and f is locally integrable on J, then the improper integrals of f and f(ϕ)ϕ converge simultaneously and are equal when convergent (Change of variable in an improper integral).

[F2]

The Gaussian integral is ex2dx=π (The Gaussian integral ex2dx=π).

Proof

technique · direct
1.1

In Γ(1/2)=0t1/2etdt, use t=u2 on proper truncations. By [F1], the improper limit is 20eu2du.

F1algebra
2.1

The integrand is even, so splitting [F2] at zero shows 20eu2du=π.

step 1.1F2algebra
3.1

Combining the two identities gives Γ(1/2)=π, with the positive square root selected because Gamma is positive.

step 2.1

Remarks

The independent Wallis-product route is Γ(1/2)=π by Wallis's product.

Depends on

Used by

Dependency tree · two levels

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Sources