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let $f(x)\in C^{1}[0,+\infty)$ , and such improper integral $$\int_{0}^{+\infty}\left(|f(x)|+|f'(x)|\right)dx$$ is convergence

show that $$\lim_{x\to+\infty}f(x)=0$$

My try: since $$(e^xf(x))'=e^x[f(x)+f'(x)]$$ but this problem is $$|f(x)|+|f'(x)|$$ so I can't.Thank you

Derek 朕會功夫
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  • Is $f$ absolutely continuous? – copper.hat Jan 06 '14 at 08:12
  • Some additional information on the subject (altough not the same) http://math.stackexchange.com/questions/108191/prove-fx-continuous-function-and-int-a-infty-fx-dx-infty-so-l – Umberto Jan 06 '14 at 08:16

1 Answers1

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Since $\int |f'| < \infty$ and $f(t) = f(t_0) + \int_{t_0}^t f'(\tau) d \tau$ we have that $f_\infty = \lim_{t \to \infty} f(t) $ exists. Since $\int|f| < \infty$, we see that we must have $f_\infty =0$.

copper.hat
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