第1个回答 2018-11-30
I = - ∫<0, +∞> xe^(-k^2x^2)d(-k^2x^2)
= - ∫<0, +∞> xde^(-k^2x^2)
= -[xe^(-k^2x^2)]<0, +∞> + ∫<0, +∞> e^(-k^2x^2)dx
= -[x/e^(k^2x^2)]<0, +∞> + (1/k) ∫<0, +∞> e^(-k^2x^2)d(kx)
(后者令 u = kx,不妨设 k > 0 )
= 0 + (1/k) ∫<0, +∞> e^(-u^2)du
= (1/k)√π/2 = √π/(2k)本回答被提问者和网友采纳