题目

te^(-pt)的反常积分怎么算的下限是0上限正无穷

提问时间：2021-04-07

答案

p≠0 时,I =∫te^(-pt)dt = -(1/p)∫tde^(-pt)
= -(1/p){te^(-pt)] -∫e^(-pt)dt]
= -(1/p){[te^(-pt)] +(1/p)[e^(-pt)]}
limte^(-pt) = limt/e^(pt)
= lim1/[pe^(pt)] = 0
则 I = -(1/p^2)[e^(-pt)] = 1/p^2