20/03/15 12:10:16 8d8gCNj7.net
>>839
I_m = ∫[0,1] (xx)(1-xx)^m dx
とおく。部分積分で
I_{m-1} - I_m = ∫[0,1] x^4･(1-xx)^{m-1} dx
= (3/2m)∫[0,1] xx･(1-xx)^m dx
= (3/2m) I_m,
I_m = {2m/(2m+3)}I_{m-1}
= ････
= {(2m)(2m-2)････2/(2m+3)(2m+1)････5}I_0
= (2m)!! / (2m+3)!! (← I_0 = 1/3)
あるいは xx=t とおいて
I_m = (1/2)Ｂ(3/2,m+1)
= (1/2)Γ(3/2)Γ(m+1)/Γ(m+5/2)
= m!(2^m) / (2m+3)!!
= (2m)!! / (2m+3)!!
m=8 のとき
16!! / 19!! = (2^15)/2078505 = 0.015765177375