>>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…