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Gamma function: ∫ dx/(1−x⁶)^(1/6) = π/3
Short answer
Put x⁶ = t, then dx = (1/6) t^{-5/6} dt. The integral becomes (1/6) B(1/6,5/6). Using Γ(1/6)Γ(5/6) = π/sin(π/6) = 2π, we obtain π/3.
Use Gamma function to prove that ∫₀¹ dx/(1−x⁶)^(1/6) = π/3.
To show: ∫₀¹ dx/(1 − x⁶)^(1/6) = π/3.
Put x⁶ = t, so x = t^(1/6) and dx = (1/6) t^(−5/6) dt. Limits are unchanged (0 to 1).
I = (1/6) ∫₀¹ t^(−5/6) (1 − t)^(−1/6) dt
= (1/6) B(1/6, 5/6) [m − 1 = −5/6 → m = 1/6; n − 1 = −1/6 → n = 5/6]
= (1/6) Γ(1/6) Γ(5/6) / Γ(1)
Since Γ(n)Γ(1 − n) = π/sin(nπ), with n = 1/6:
Γ(1/6)Γ(5/6) = π / sin(π/6) = π/(1/2) = 2π.
I = (1/6)(2π) = π/3. Proved.
✓ Answer checked by Claude Opus 5.5 (high effort)
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