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Is $\int_0^{+\infty}x^2 \sin(\frac{\cos x^3}{x+1})$ convergent, absolutely convergent?

I tried to use equivalence to $\sin(\frac{\cos x^3}{x+1})$ - $\frac{\cos x^3}{x+1}$. But i cannot use this method, cause this function isn't > 0 on all of the domain of integration.

Nigel Overmars
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Eugene Korotkov
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  • you should begin by putting $t=x^3$ – hamam_Abdallah May 10 '17 at 23:01
  • You can still use this method (although you should justify ignoring the lower order terms in the Taylor expansion). After that the substitution $u=x^3$ will be helpful, and then you can use integration by parts. It should be fairly obvious that the integral does not converge absolutely. – Jason May 10 '17 at 23:02

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