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=∫0f(x)dx,求级数的和.

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答案an=∫0f(x)dx=∫0e-xsin nxdx=-∫0sin nxd(e-x) =-(e-xsin nx|0-n∫0e-xcos nxdx) =-n∫0cos nxd(e-x) =-n(e-xcos nx|0+n∫0e-xsin nxdx) =-n(e-2π-1)-n20e-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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