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.
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.