上行处: ∫<0,+∞>ue^(-u)du = - ∫<0,+∞>ude^(-u)
= - [ue^(-u)] <0,+∞> + ∫<0,+∞>e^(-u)du
= 0 - [e^(-u)] <0,+∞> = 0 - 0 + 1 = 1,
此处用了: limue^(-u) = limu/e^u (∞/∞)
= lim1/e^u = 0.
下行处:因 L(θ) = (2^n x1x2...xn/θ^n)e^[-(1/θ)∑(xi)^2]
则 lnL(θ) = (nln2 + lnx1 + lnx2 + ... + lnxn - nlnθ) [-(1/θ)∑(xi)^2]
= (nln2 + ∑lnxi - nlnθ) [-(1/θ)∑(xi)^2], 即得。