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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

jontao
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  • 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. – Paul Sinclair Jul 10 '23 at 21:47
  • If you have $\int_1^{\infty} |g(x)| dx < +\infty$ then $\int_1^{\infty} g(x) dx $ is finite.(Absolute Convergence Criteria) – jontao Jul 11 '23 at 08:27
  • 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 $G$. – Paul Sinclair Jul 11 '23 at 12:16
  • 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? – jontao Jul 11 '23 at 14:25
  • 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. – Paul Sinclair Jul 11 '23 at 17:11

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