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?

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

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.