How to prove $ \int_{1}^{ \infty} \frac{1}{ (1+x^3)^3 } dx$ is convergent?

Question

I am trying to determine whether the following improper integral is convergent or not.

$$ \int_{1}^{ \infty} \frac{1}{ (1+x^3)^3 } dx$$

I tried the following: $l = \lim_{x \to a} ((x-a)^k)f(x)$, if $l \in [0, \infty)$ and $k < 1$, then the integral is convergent. But I can't use it well. Can someone help me? Thank you.

Answer 1

If $x \in [1,\infty)$, then $\frac{1}{(1+x^3)^3} < \frac{1}{x^6}<\frac{1}{x^2}$. So for every $n \in \Bbb N$, we have that $\int_{1}^n \frac{1}{(1+x^3)^3}<\int_{1}^n\frac{1}{x^2}$. But $\int_{1}^n\frac{1}{x^2} = 1 - \frac{1}{n}$. Therefore

\begin{equation} \int_{1}^n \frac{1}{(1+x^3)^3}<1-\frac{1}{n} \end{equation}

Now taking limits we get

\begin{equation} \int_{1}^{\infty} \frac{1}{(1+x^3)^3}= \lim_{n \to \infty}\int_{1}^n \frac{1}{(1+x^3)^3} < \lim_{n \to \infty}\left( 1-\frac{1}{n} \right) = 1 \end{equation}

This shows the improper integral converges.