# Does the integral $\int_{2}^{\infty}\frac{dx}{\sqrt{1+x^{3}}}$ exists?

I need to prove this. I need your help to verify that my proof is correct (or not) please.

Prove that this integral exists: \begin{align} \int_{2}^{\infty}\frac{dx}{\sqrt{1+x^{3}}} \end{align}

My attempt:

Fist we need to observe that $$\frac{1}{\sqrt{1+x^{3}}}<\frac{1}{\sqrt{x^{3}}}\Longrightarrow \int_{2}^{\infty}\frac{dx}{\sqrt{1+x^{3}}}<\int_{2}^{\infty}\frac{dx}{\sqrt{x^{3}}}$$

Next, we note that $$\lim_{x \rightarrow \infty} \frac{\frac{1}{\sqrt{1+x^{3}}}}{\frac{1}{\sqrt{x^{3}}}}=1 \Longrightarrow$$ for $$f(x)=\frac{1}{\sqrt{1+x^{3}}}$$, $$g(x)=\frac{1}{\sqrt{x^{3}}}$$ both integrals converges or both diverges.

By last, integrals of the form $$\int_{1}^{\infty}\frac{dx}{x^{p}}$$ converges if $$p>1$$, $$\Longrightarrow \int_{1}^{\infty}\frac{dx}{\sqrt{x^{3}}}$$ converges $$\Longrightarrow \int_{2}^{\infty}\frac{dx}{\sqrt{x^{3}}}$$ converges

That implies that, $$\int_{2}^{\infty}\frac{dx}{\sqrt{1+x^{3}}}$$ converges, therefore it exists.

Is it correct? Is there another way to prove it? Thank you very much

• That is correct but once you improve your proof writing these proofs can be written much shorter. – Derek Luna Jan 10 at 22:19

Your solution is correct, though you didn't really need to use the limit comparison test. You could just stop after the first line. Since $$\frac{1}{\sqrt{1+x^3}}\leq\frac{1}{\sqrt{x^3}}$$ and the integral $$\int_2^{\infty}\frac{1}{\sqrt{x^3}}dx$$ converges, we know from the usual comparison test that $$\int_2^{\infty}\frac{1}{\sqrt{1+x^3}}dx$$ converges as well. Of course it is important to note that these are nonnegative functions in the solution.
$$0\leq \int_2^{+\infty} \frac{1}{\sqrt{1+x^3}} dx \leq \int_2^{+\infty} \frac{1}{x^{3/2}} dx$$ and the last integral is finite. Therefore also the one in the middle of the inequalities is finite.