题目

求函数∫(0→x)sint/tdt关于x的幂级数

提问时间：2021-01-28

答案

[ ∫(0→x)sint/tdt ]'=sinx/xsinx=x-(1/3!)x³+(1/5!)x^5-(1/7!)x^7+...=Σ(-1)^n(1/(2n+1)!)x^(2n+1) n=0→∞sinx/x=1-(1/3!)x²+(1/5!)x^4-(1/7!)x^6+...=Σ(-1)^n(1/(2n+1)!)x^(2n) n=0→∞上式积分后得...