- 515 名前：１３２人目の素数さん mailto:sage [2009/10/04(日) 09:41:00 ]
- >>376, >>502(7)
1/(t^p + 1) = x とおくと、
t = (1/x - 1)^(1/p),
p･dt = (-1/x^2)(1/x - 1)^(1/p - 1) dx,
(左辺) = ∫[0,1] (1/x)(1/x -1)^(1/p -1) dx
= ∫[0,1] x^((1 -1/p)-1) (1-x)^(1/p -1) dx
= Ｂ(1 -1/p, 1/p)
= Γ(1 -1/p)Γ(1/p) / Γ(1)
= π/sin(π/p),
等式の希ガス…
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