Question

判断瑕积分的敛散性 ∫1/(3次根号下(x^2*(1-x))) dx 积分上限是1 下限是0

判断瑕积分的敛散性 ∫1/(3次根号下(x^2*(1-x))) dx 积分上限是1 下限是0

Answer 1

∵x=0与x=1是原积分的两个瑕点
∴把它分成两个积分判断,即
原积分=∫(0,1)dx/(x²(1-x))^(1/3) (∫(0,1)表示从0到1积分,以下类同)
=∫(0,1/2)dx/(x²(1-x))^(1/3)+∫(1/2,1)dx/(x²(1-x))^(1/3)
设f(x)=1/(x²(1-x))^(1/3)
∵lim(x->0+)(x^(2/3)*f(x))=lim(x->0+)(1/(1-x)^(1/3))=1
∴积分∫(0,1/2)dx/(x²(1-x))^(1/3)收敛
∵lim(x->1-)((1-x)^(1/3)*f(x))=lim(x->1-)(1/x^(2/3))=1
∴积分∫(1/2,1)dx/(x²(1-x))^(1/3)收敛
故原积分收敛.