Question

求x^2/w^2*[e^(-x^2/(2*w^2))]在x>0时的积分,其中的w是常数,对x积分

Answer 1

令x/w√2=t ,代入得：

原式=2w√2∫(0,+∞)t^2e^(-t^2)dt

=-w√2∫(0,+∞)tde^(-t^2)

=-w√2(te^(-t^2)|(0,+∞) -∫(0,+∞)e^(-t^2)dt)

=w√2∫(0,+∞)e^(-t^2)dt

=w√2*√π/2

=w√(π/2)