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Consider $f:[1,\infty)\to \mathbb{R}$ of class $C^1$ such that $ \int_{1}^{+\infty} |f'(x)| dx $ is finite. Prove that $I = \int_{1}^{+\infty} f(x) dx $ exist iff exist $\lim_{n \to +\infty} \int_{1}^{n} f(x) dx$ ($n$ is an integer).

I know for hypothesis that $ \int_{1}^{+\infty} f'(x) dx = \lim_{x \to +\infty} f(x) - f(1)$ is finite so $L = \lim_{x \to +\infty} f(x)$ is finite. Now $L > 0 \to \int_{1}^{+\infty} f(x) dx = +\infty$ and $L < 0 \to \int_{1}^{+\infty} f(x) dx = -\infty$ but I don't know how to continue

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  • $\begingroup$ How do you know $\int_1^\infty f'(x)\,dx$ is finite? You were told the integral of the absolute value is finite, not the function itself. $\endgroup$ Jul 10, 2023 at 21:47
  • $\begingroup$ If you have $\int_1^{\infty} |g(x)| dx < +\infty$ then $\int_1^{\infty} g(x) dx $ is finite.(Absolute Convergence Criteria) $\endgroup$
    – jontao
    Jul 11, 2023 at 8:27
  • $\begingroup$ Okay - you did know that much. What you don't seem to realize is that the question amounts to giving you a function $$G: [1,\infty)\to \Bbb R : x \mapsto \int_1^xf(x)dx$$ and asking you to prove that $\lim_{x\to\infty} G(x)$ and $\lim_{n\to\infty}G(n)$ converge or diverge together. That is, the behavior of $G$ is such that its behavior on integers around $\infty$ controls its behavior over all values around $\infty$. This is what you need to examine. How does the information about $f$ restrict $G$. $\endgroup$ Jul 11, 2023 at 12:16
  • $\begingroup$ Thank you, but I don't understand how could I link limts between integers and reals, maybe there is a link I don't know, can you give me a hint? $\endgroup$
    – jontao
    Jul 11, 2023 at 14:25
  • $\begingroup$ That the limit over the continuum converging implies the limit on integers converging is almost trivial. So you need to concentrate on showing that if the limit on integers converges, then because of the given properties of $f$, it must eventually be that $f$ cannot go too wild between the integers (at low integers, yes, but as you approach $\infty$, it must eventually settle down), and therefore the behavior at the integers is sufficient to imply the whole thing. $\endgroup$ Jul 11, 2023 at 17:11

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