吉米多维奇数学分析习题集1925题.doc

《吉米多维奇数学分析习题集》1925题
=arctg ?sinπ?cosπ?ln(x2?2x cosπ+1)+C
解：∵　x2n+1=x2n?ei(2N+1)π=x2n?(eiπ)2n，则x2n+1必有因式x?eiπ，N为任意整《吉米多维奇数学分析习题集》1925题
=arctg ?sinπ?cosπ?ln(x2?2x cosπ+1)+C
解：∵　x2n+1=x2n?ei(2N+1)π=x2n?(eiπ)2n，则x2n+1必有因式x?eiπ，N为任意整数．
∴　x2n+1的因式集为{x?eiπ|k=1,2,…,2n.}，取a=eiπ，
x2n+1=(x?eiπ)=(x?a2k?1)=(x?a1)(x?a3)…(x?a2k?1)…(x?a4n?1)．
则　x2n+1可表述为：x2n+1=(x?a2k1?1)(x?a2k2?1)…(x?a2kl?1)…(x?a2k2n?1)，(2k?1)=(2kl?1)．
∴　x2n+1的x2n?l项系数为：(?a2k1?1)(?a2k2?1)…(?a2kl?1)
=(?1)l?a2k1?1?a2k2?1?…?a2k2n?1=(?1)l? a2kj?1=0．
x2n+1的常数项为：(?eiπ)=(?a2kl?1)=a2k?1=a2k?1?a?(2k?1)=1=1．
∴　 　=  = ?(x?a1)(x?a3)…(x?a2k?1)…(x?a4n?1)
=(x?a2k1?1)(x?a2k2?1)…(0?a2kl?1)…(x?a2k2n?1)=??l．
??l=(x?a2k1?1)(x?a2k2?1)…(0?a2kl?1)，
∴　??l的x2n?l项系数为：(?a2k1?1)(?a2k2?1)…(?a2kl?1) =0．
??l的常数项为：(?eiπ)=1．
∴　 =??l=1=．
则　= =(+)=．
∴　=
=arctg ?sinπ?cosπ?ln(x2?2x cosπ+1)+C．