18/07/23 16:11:29.41 4rZpnird.net
>>226
J_m = ∫[0,1] 1/(1+xx)^m dx とおくと I_n = J_{2n},
2m(J_{m+1} - J_m) = -2m∫[0,1] xx/(1+xx)^{m+1} dx
= [ x/(1+xx)^m ](x=0,1) - J_m
= 1/2^m - J_m,
J_{m+1} = {(2m-1)/2m}J_m + 1/(2m･2^m),
また J_0 =1，J_1=π/4 ゆえ真
例
J_2 = (π+2)/8,
J_3 = (3π+8)/32,
J_4 = (15π+44)/192,