设f(x)在区间(0,+∞)内绝对可积,则∫0+∞f(x)sinnxdx=0.

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问题 设f(x)在区间(0,+∞)内绝对可积,则0+∞f(x)sinnxdx=0.

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答案由于f(x)在(0,+∞)内绝对可积,故对任意的ε>0,存在N>0,当a≥N时,有 |∫a+∞f(x)dx|≤∫0+∞|f(x)|dx<ε/2, 又由于[*]∫0Nf(x)sinnxdx=0.故存在M>0,当n>M时,有|∫0Nf(x)sinnxdx|<ε/2,从而当n>M时,有 |∫0+∞f(x)sinnxdx|≤|∫0Nf(x)sinnxdx|+|∫N+∞f(x)sinnxdx| <[*]+|∫N+∞|f(x)|sinnxdx|<[*]=ε. 因此,[*]|∫0+∞f(x)sinnxdx=0.

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