∫ 1/√( x^3-1 ) dx ( x = 1〜∞ )
= ( 1/√3 )B( 1/3 , 1/2 )
⇔
B( 1/3 , 1/2 )
= √3∫ 1/√( x^3-1 ) dx ( x = 1〜∞ )
x^3-1
= ( x-e1 )( x-e2 )( x-e3 )
= ( x-1 )( x-ζ3 )( x-ζ3^2 ) ( ζ3 = cos( 2π/3 )+√-1sin( 2π/3 ) )
k^2 = 1-k’^2
= ( e2-e3 )/( e1-e3 )
= -ζ3
k’2
= 1+ζ3^2
= -ζ3^2
k’
= √-1ζ3
= ζ12^( -1 )
( 1/2 )∫ 1/√( x^3-1 ) dx ( x = 1〜∞ )
= K( k )/√( e1-e3 )
= ( 1/A.G.M.( 1 , k’ ) )( π/2 )/√( e1-e3 )
⇒
∫ 1/√( x^3-1 ) dx ( x = 1〜∞ )
= ( π/A.G.M.( 1 , k’ ) )/√( e1-e3 )
⇒
B( 1/3 , 1/2 ) = 2∫ ( sinθ )^( -1/3 ) dθ ( θ = 0〜π/2 )
= √( 3/( e1-e3 ) )( π/A.G.M.( 1 , k’ ) )
= √( 3/( √3ζ12 )( π/A.G.M.( 1 , ζ12^( -1 ) ) )
= 3^( 1/4 )( π/( ζ24A.G.M.( 1 , ζ24^( -2 ) ) )
= 3^( 1/4 )( π/A.G.M.( ζ24 , ζ24^( -1 ) ) )
= 3^( 1/4 )( π/A.G.M.( 1 , cos( π/12 ) ) )
やっと出ました・・・・・・・
Remark :
B( 1/4 , 1/2 ) = 2ω = 2( π/A.G.M.( 1 , √2 ) )
= 4^( 1/4 )( π/A.G.M.( 1 , cos( π/4 ) ) )