Question

怎么求广义积分

Answer 1

∫(1/x²)[sin(1/x)]dx
=-∫[sin(1/x)]d(-1/x)
=∫[sin(1/x)]d(1/x)
=-cos(1/x)
当x→+无穷时，1/x→0+，-cos(1/x)→-1
当x→2/π时，1/x→π/2，-cos(1/x)→0
所以原广义积分=-1-0=-1

Answer 2

let

u=1/x
du = -(1/x^2) dx
x=2/π, u=π/2
x=+∞, u=0
∫(2/π->+∞) (1/x^2)sin(1/x) dx
=-∫(π/2->0) sinu du
= [cosu]|(π/2->0)
=1