换元法, 令 \lambda x=t
令 λx = t， 则 x = t/λ， dx = dt/λ,
原积分 I = ∫<0, +∞> t^2e^(-t)dt/λ = (1/λ) ∫<0, +∞> t^2e^(-t)dt
= (-1/λ) ∫<0, +∞> t^2de^(-t)
= (-1/λ){[t^2/e^t]<0, +∞> - 2∫<0, +∞> te^(-t)dt}
= (-1/λ)[0 + 2∫<0, +∞> tde^(-t)]
= (-2/λ){[t/e^t]<0, +∞> - ∫<0, +∞>e^(-t)dt}
= (-2/λ){0 +[e^(-t)]<0, +∞>} = 2/λ
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