反常积分求定积分上2下0,dx/(1-x)^2.答案是发散,

Question

反常积分
求定积分上2下0,dx/(1-x)^2.答案是发散,

Answer 1

分成两个区间分别积分,[0,1-],[1+,2],
原式=lim [t→1-]∫ [0,t]dx/(1-x)^2+lim [t→1+]∫ [t,2]dx/(1-x)^2
=lim [t→1-](-1)∫ [0,t]d(1-x)/(1-x)^2+lim [t→1+](-1)∫ [0,t]d(1-x)/(1-x)^2
=-(1-x)^(-2+1)/(-2+1)[0,1-]+[-(1-x)^(-2+1)/(-2+1)[1+,2]
=1/(1-x)[0,1-]+1/(1-x)[1+,2]
=∝
∴发散.