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I need help with this improper integral:

$$\int_1^\infty\ln x\cdot \arcsin\left(\frac{1}{x^2}\right)\,\mathrm dx $$

I need to determine whether it is convergent or divergent.

I can only use the Limit Comparison Test, and the limit: $\lim_{x \to 0} \frac{\arcsin(x)}{x} = 1$.

I had tried again and again to find some function of the form $\frac{1}{x^p}$ or $\frac{1}{x\ln^px}$ to compare with the function in the integral, but could not find anything that would work. I'm sure that I am missing something obvious.

Help will be much appreciated.

Thanks.

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  • $\begingroup$ Uh... well, you know that $\lim_{x\to0}\frac{\arcsin(x)}x=1$, so you can deduce that $\lim_{x\to\infty}\frac{\arcsin(1/x^2)}{1/x^2}=1$, so... $\endgroup$
    – Simply Beautiful Art
    Commented Jul 10, 2017 at 22:00
  • $\begingroup$ Of course, I've noticed that :) Didn't manage to get somewhere from there though. As I have said, I'm sure I'm missing something obvious, but it is not that. $\endgroup$
    – tfreifeld
    Commented Jul 10, 2017 at 22:04
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    $\begingroup$ Could you show what happened after you've noticed that? Showing your work helps us help you, but if you don't include you're work, we can't know what you've tried or where you went wrong. $\endgroup$
    – Simply Beautiful Art
    Commented Jul 10, 2017 at 22:05
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    $\begingroup$ You can't use the plain old comparison test? $\endgroup$
    – zhw.
    Commented Jul 10, 2017 at 22:16
  • $\begingroup$ @SimplyBeautifulArt I tried different things of the form I have mentioned in the question. Didn't manage to get rid of the $\ln$ $\endgroup$
    – tfreifeld
    Commented Jul 10, 2017 at 22:17

1 Answer 1

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I believe I got it right eventually.

I've used to Limit Comparison Test with the function $\frac{1}{x\cdot ln^2x}$, for which I know the integral $\int_1^\infty$ is convergent.

From there, I got to:

$$\lim_{x \to \infty} \frac{\ln ^3 x}{x} \cdot \frac{\arcsin (\frac{1}{x^2})}{\frac{1}{x^2}}$$

I know $\lim_{x \to \infty} \frac{\ln ^3 x}{x} = 0$ and that $\lim_{x \to \infty} \frac{\arcsin (\frac{1}{x^2})}{\frac{1}{x^2}} = 1$, so the entire thing goes to zero. According to a "special" case of the Limit Comparison Test, if $\lim_{x \to \infty} \frac{f(x)}{g(x)} = 0$, and $g(x)$ is convergent, then $f(x)$ is convergent as well. Therefore, the original integral in the question is convergent.

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