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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

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

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Yes, you compute a limit that is not necessary to compute. You just need to bound your integrand by something that is integrable (like you did):

$$ 0\leq \int_2^{+\infty} \frac{1}{\sqrt{1+x^3}} dx \leq \int_2^{+\infty} \frac{1}{x^{3/2}} dx $$ and the last integral is finite. Therefore also the one in the middle of the inequalities is finite.

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