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Assume I have a function $ f(x) $ and I want to know if the improper integral $ \intop_{a}^{b}f\left(x\right)dx $ converge/diverge.

Can I use the limit comparison test with a riemann integrable function $ g $ ?

For example, I want to check if the integral $ \intop_{0}^{1}\frac{\sin^{2}\left(x^{3}\right)}{x^{6}}dx $ converge.

So can I use the limit comparison test with $ x^3 $ ? because that will lead me to:

$ \frac{\frac{\sin^{2}\left(x^{3}\right)}{x^{2}}}{x^{3}}=\frac{\sin^{2}\left(x^{3}\right)}{x^{6}}=\left(\frac{\sin\left(x^{3}\right)}{x^{3}}\right)^{2}\underset{x\to0}{\longrightarrow}1 $

Well, actually the "improper" integral that I gave as example is also riemann integrable since that limit exists at $ x=0 $. But thoretically speaking, will this work ? is it legit?

Thanks in advance.

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Riemann improper integrals are defined for 2 cases: when function is unbounded and when integration area is unbounded. In your case, as you have checked it, your function is continuous on $(0,1]$ and have limit in $0^+$, so it can be continued on $[0,1]$ continuously, therefore is even uniformly continuous and bounded. Integral from it is usual integral, not improper.

As for comparison test, then, for example for unbounded integration area, then one example is following:

If $0\leqslant g(x) \leqslant f(x)$ on some interval $[a, \infty)$, $f,g$ integrable on each $[a,b]$, then if $\int\limits_{a}^{\infty}f(x)\,dx$ converges, so does $\int\limits_{a}^{\infty}g(x)\,dx$. If last diverges diverges, so does former.

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