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I want to determine the convergence of the following improper integral:$$\int_0^{\infty} \frac{1}{1+x^4\sin^2x}\,dx.$$

I have tried comparison tests for the same. $$\int_0^{\infty} \frac{1}{1+x^4}\,dx$$ comes out to be convergent while $$\int_0^{\infty} \frac{1}{1-x^4}\,dx$$ comes out to be divergent. Thus we can't say actually anything about the convergence of this integral from these two comparisons. How can I proceed from here? Please suggest.

DMcMor
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HARVEER RAWAT
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    $0\le \sin^2 x \le 1$ – V.G Mar 02 '21 at 13:48
  • @ClaudeLeibovici: No, I can't. I wrote it because OP was comparing it with $\dfrac{1}{1-x^4}$ where he was considering $-1\le \sin^2 x$, so that's why I wrote that. I even wrote an answer based on this inequality but later realised $\le1$ also doesn't make it convergent...although I have found some posts related: this for example. – V.G Mar 02 '21 at 14:05
  • @LightYagami. The link you give is very interesting, for sure. Thank you. Cheers :-) – Claude Leibovici Mar 02 '21 at 14:08
  • @ClaudeLeibovici: You may check the new link also... it is better than that. – V.G Mar 02 '21 at 14:15
  • @LightYagami. This one is more than great ! Thanks again. – Claude Leibovici Mar 02 '21 at 14:19

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