∫0到-∞ 1/e^x+e^-xdx 反常积分值

Question

如题所述

Answer 1

∫1/[e^x+e^(-x)] dx
=∫e^x/[(e^x)^2+1] dx
=∫1/[(e^x)^2+1] d(e^x)
=arctan(e^x)+C
当x=0时，arctan(e^x)=arctan(e^0)=arctan1=π/4；
当x=-∞时，arctan(e^x)=arctan[e^(-∞)]=arctan0=0

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