判别广义积分∫(上e 下1⼀e) ln|x-1|⼀(x-1) dx的敛散性

2025-06-27 14:22:51
推荐回答(1个)
回答1:

这个积分应该是收敛的;
∫{x=1/e→e} [ln|x-1|/(x-1)]dx
=∫{x=1/e→1-δ} [ln(1-x) /(x-1)] dx + ∫{x=1-δ→e} [ln(x-1) /(x-1)] dx ……δ→0
=(1/2)ln²(1-x)|{1/e,1-δ}+(1/2)ln²(x-1)|{1+δ,e}……δ→0
=ln²δ-(1/2)ln²[1-(1/e)]+(1/2)ln²(e-1)-ln²δ……δ→0
=(1/2)ln²{(e-1)/[(e-1)/e]}=1/2