记 S(x) = Σ (n+1)(x-1)^n, 则∫<1,x> S(t)dt = ∫<1,x> (n+1)(t-1)^ndt = [(t-1)^(n+1)]<1,x> = (x-1)^(n+1) = (x-1)^2+(x-1)^3+(x-1)^4+......= (x-1)^2/[1-(x-1)] = (x-1)^2/(2-x).收敛域 -1于是 S(x)=[(x-1)^2/(2-x)]' = (x-1)(3-x)/(2-x)^2. 收敛域 0
