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Find all the values $\alpha \in(0,\infty)$ such that the improper integral $$\int\limits_0^\infty \frac{\Bbb dx}{1+x^{\alpha}\sin^2x}$$ is convergent.

My attempt is to analyze the cases (i) $\alpha =1$, (ii) $\alpha >1$, and $\alpha <1$.

The case (i) gives divergence. I don’t know how to handle the other two cases. Does anyone have any suggestions for these cases?

Rócherz
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Harry
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  • Can you write out your work with $\alpha=1$? – ShyamalSayak Nov 04 '23 at 00:56
  • well, for case $\alpha=1$, we have $\int\limits_0^\infty \dfrac{dx}{1+x\sin^2x} \geq \int\limits_0^\infty \dfrac{dx}{1+x}$. Because $\int\limits_0^\infty \dfrac{dx}{1+x}$ diverges, it means that for case $\alpha =1$, the improper integral diverges, right? – Harry Nov 04 '23 at 01:02
  • Do you think that you can use the same idea for the rest? – ShyamalSayak Nov 04 '23 at 01:05
  • I think it can be reused to finish the case $\alpha<1$ also (hence in this case the improper integral diverges). But for $\alpha>1$, I don't think so, because $\int\limits_0^\infty\dfrac{dx}{1+x^\alpha}$ converges. – Harry Nov 04 '23 at 01:09
  • Can you compare your integrand with $1/(x^{\alpha}\sin ^2x)$, and then possibly use by parts? – ShyamalSayak Nov 04 '23 at 01:17
  • hmm, I don't think so – Harry Nov 04 '23 at 01:23

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