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$f :[1, \infty)\rightarrow\mathbb{R}$ is a continuous function, with a bounded antiderivative $F$

It follows that, the improper integral $$\int_{1}^{\infty} \frac{f(x)}{x^{\alpha}} dx$$ exists for all $\alpha \in \mathbb{R^{+}}$

I don't want a full solution, just an explanation or a hint. Like, what can I deduce from a function having a bounded $F$?

Another question, is there an $\epsilon-\delta$ definition of convergence for improper integrals?

yarafoudah
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  • The definition of $\int_1^\infty$ existing is $\lim_{t \to \infty} \int_1^t$ existing. The $\varepsilon$-$\delta$ definition follows from that. – Theo Bendit Nov 20 '17 at 13:10
  • Try replacing $f$ with $F'$, and applying integration by parts. – Theo Bendit Nov 20 '17 at 13:14
  • I tried it. I get to the integral $\int_{1}^{\infty} F(x) \ x^{(-\alpha - 1)} dx$. If I integrate again by parts, it will all cancel each other out. – yarafoudah Nov 20 '17 at 14:00
  • Now you can use the fact that $F$ is bounded. So, this integral is absolutely bounded by a constant times the integral $\int_1^\infty x^{-\alpha - 1}\mathrm{d}x$, which converges. – Theo Bendit Nov 20 '17 at 14:21
  • Yes, I get it now. Thank you! :) – yarafoudah Nov 20 '17 at 14:49

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