2. 考研
  3. 设f(x)满足f’(x)+f(x)=ne-xcos nx，n为正整数，f(0)=0． 设an=∫02πf(x)dx，求级数的和．

设f(x)满足f’(x)+f(x)=ne-xcos nx，n为正整数，f(0)=0． 设an=∫02πf(x)dx，求级数的和．

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问题 设f(x)满足f’(x)+f(x)=ne-xcos nx，n为正整数，f(0)=0．
设an=∫02πf(x)dx，求级数的和．
选项
答案an=∫02πf(x)dx=∫02πe-xsin nxdx=-∫02πsin nxd(e-x) =-(e-xsin nx｜02π-n∫02πe-xcos nxdx) =-n∫02πcos nxd(e-x) =-n(e-xcos nx｜02π+n∫02πe-xsin nxdx) =-n(e-2π-1)-n2∫02πe-xsin nxdx =-n(e-2π-1)-n2an． 移项，得an=[*]．故 [*] 两边同时积分，得∫0xS’1(t)dt=∫0x[*]dt，即 S1(x)-S1(0)=ln(1+x)，S1(0)=0， 故S1(x)=In(1+x)．令 [*]
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