Question

反常积分∫(上积分为+∞,下积分为-∞)1/1+x^2dx等于

Answer 1

∫(-∞,∞) 1/(1+x²) dx

= ∫(-∞,0) 1/(1+x²) dx + ∫(0,∞) 1/(1+x²) dx

= [arctanx](-∞,0) + [arctanx](0,∞)

= [0-(-π/2)] + [π/2-0]

= π