广义积分1到+∞dx/x根号(x^2-1)

2024-11-02 06:41:10
推荐回答(1个)
回答1:

令x=ch u,则dx= shu du

则∫(1到+∞) dx/x根号(x^2-1)
=∫(0到+∞) shu du/(chu ·shu)
=∫(0到+∞) du/chu
=2∫(0到+∞) du/(e^u+e^(-u))
=2∫(0到+∞) e^u du/((e^u)²+1)
=2∫(0到+∞) d(e^u)/((e^u)²+1)
=2 arctan(e^u) |(0到+∞)
=2 [lim(u→+∞)arctan(e^u) - arctan(e^0)]
=2 (π/2 - π/4)
=π/2