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TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Euler's reflection formula

Statement

For every zCZ,

Γ(z)Γ(1z)=πsin(πz).

The identity extends meromorphically to all zC.

Facts & Assumptions

Given: The reciprocal-Gamma product and the sine product.

[L1]

Reciprocal Gamma has the product 1/Γ(z)=zeγzn1(1+z/n)ez/n (The Weierstrass product for reciprocal Gamma).

[L2]

Sine has the product sin(πz)=πzn1(1z2/n2) (The Weierstrass product for sine).

[L3]

Harmonic numbers satisfy Hn=logn+γ+o(1) (The Euler–Mascheroni constant and the harmonic asymptotic).

Proof

technique · direct
1.1

Apply [L1] at 1z. For the Nth partial product, 1Γ(1z)=(1z)eγ(1z)limNn=1N(1+1zn)e(1z)/n. The identity (1z)n=1N(1+1zn)=(N+1)n=1N+1(1zn) therefore gives 1Γ(1z)=eγzlimN(N+1)eγHNez/(N+1)n=1N+1(1zn)ez/n. By [L3], the scalar prefactor tends to 1, so 1Γ(1z)=eγzn1(1zn)ez/n.

L1L3algebra
2.1

Multiplying step 1.1 by the product for 1/Γ(z) from [L1], the exponential factors cancel and one gets 1Γ(z)Γ(1z)=zn1(1z2n2). By [L2], the product on the right is sin(πz)/π. Therefore 1Γ(z)Γ(1z)=sin(πz)π on CZ, which is equivalent to the displayed formula.

step 1.1L1L2algebra

Depends on

Used by

Dependency tree · two levels

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Sources