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I have this exercise:

Determine if the following improper integrals exist:

  • $$\int_{0}^{\infty} \frac{1}{\sqrt{1+x^3}}dx$$
  • $$\int_{0}^{\infty} \frac{x}{1+x^{\frac{3}{2}}}dx$$
  • $$\int_{0}^{\infty} \frac{1}{x\sqrt{1+x}}dx$$

I'm lost and I don't know how should I do it. I know that $$\int_{0}^{\infty} f(x)dx = \lim_{n \to \infty} \int_{0}^{n}f(x)dx$$

And so, I try to calculate the integrals from zero to n, but I wasn't able to obtain an expression for them. Then I compute the integrals using Mathematica and those integrals seems to be really hard to compute manually, at least the first one, because the second and third integrals diverge.

I understand that the exercise only asks for a criteria for existance and that I don't need to compute an expression for the integrals, just give an argument of why do I say the integral exists or not, but I'm really stuck.

Is there a general result for improper integral's existence that I'm not using?

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  • $\begingroup$ Do you know things like the comparison theorem? $\endgroup$ – Randall Oct 12 '18 at 3:21
  • $\begingroup$ Find an equivalent of the integrated functions to deal with integrals of the functions $x \mapsto x^\alpha$, which are known cases. $\endgroup$ – AlexL Oct 12 '18 at 3:22
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Result:

If $\lim_{x \rightarrow \infty} x^p f(x)=A < \infty$, then $$\int_a^\infty f(x) \;dx \;\text{is} \begin{cases} < \infty & \text{if}\;p>1\\\\\text{diverges}& \text{otherwise}\;\end{cases} $$

For the first bullet, consider $$\lim_{x \rightarrow \infty}x^{3/2}f(x)=\lim_{x \rightarrow \infty}x^{3/2}\frac{1}{x^{3/2}(1+1/x^3)^{1/2}}=\lim_{x \rightarrow \infty}\frac{1}{(1+1/x^3)^{1/2}}=1< \infty$$ so this improper integral exist!

For the second bullet, consider $\lim x^{1/2}f(x)$ and for the third case, consider $\lim xf(x)$

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    $\begingroup$ +1 nice. It's better to give a link of this lemma (result). $\endgroup$ – Nosrati Oct 12 '18 at 10:10
  • $\begingroup$ Thank you @Chinnapparaj one final question: is there a link or reference where I could read this theorem/lemma/result and it's proof? $\endgroup$ – frl93 Oct 12 '18 at 19:44
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    $\begingroup$ Yea! See Theorem $1$ (page no. 4 of pdf (or) page no. 324 of the material) in this paper $\endgroup$ – Chinnapparaj R Oct 13 '18 at 2:00
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I choose the third, with $x+1=u$ $$\int_{0}^{\infty} \frac{1}{x\sqrt{1+x}}dx=\int_{1}^{\infty} \frac{1}{(u-1)\sqrt{u}}du > \int_{1}^{\infty} \frac{1}{u\sqrt{u}}du=\infty$$

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  • $\begingroup$ Thank you, you used the comparision theorem right? $\endgroup$ – frl93 Oct 12 '18 at 19:45

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