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Let $f(x)$ be a polynomial of degree $k\geq 2$, such that $f(x)>0$ for all $x>a$. Prove that $\lim_{x\to\infty}\int_{a}^x \frac{1}{f(x)}\,dx<\infty$.

The intuition is clear, we must somehow use that $\int_a^\infty \frac{1}{x^2}\,dx<\infty$. But I am trying to find and prove some inequality involving $f(x)$ to use this fact. I thought of proving either that there exists $c$ such that $f(x)\geq cx^2$ for $x$ big enough or $f(x)\geq \frac{1}{k}(x-a)^k$ for all $x>a$ using some uniform continuity argument.

Could someone help?

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  • $\begingroup$ Consider $f(x)=x^2.$ Is $\int_0^{\infty} \frac{dx}{x^2}<\infty?$ $\endgroup$ – mfl Oct 15 '18 at 13:56
  • $\begingroup$ @mfl no, but $\int_a^{\infty} \frac{dx}{x^2}<\infty$ for $a>0$. $\endgroup$ – Heinz Doofenschmirtz Oct 15 '18 at 14:03
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For simplicity we can suppose that $a>0$.

As you have said proving that the existence of $c>0$ such that $f(x) \geq c x^2$ $\forall x>a$ is sufficient to conclude.

Let us consider: $$g(x) =\frac{x^2}{f(x)}$$

defined on $(a,+\infty)$. Then:

  • $g$ is well defined
  • $g$ is continuous
  • as $\deg(f) \geq 2$ there exits $l \in \mathbb{R}$ such that $\lim_{x \to \infty} g(x)=l$

so by standard arguments you can conclude that $g$ is bounded, so there exists $c>0$ such that: $$g(x) =\frac{x^2}{f(x)} \geq c$$


For a general $a$, instead of $x^2$ it is a better idea to consider $(x-a+1)^2$ to avoid problems.

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