求区间（0,x）上∫sint/tdt在x=0处的幂级数展开式，并确定它收敛于该函数的区间

Question

如题所述

Answer 1

f(x) = ∫<0,x> sintdt/t = ∫<0,x> sintdt/t
= ∫<0,x>∑<n=0,∞>(-1)^n*t^2ndt/(2n+1)!
= ∑<n=0,∞>(-1)^n*x^(2n+1)/[(2n+1)(2n+1)!]
(-∞<x<+∞)本回答被提问者采纳