$\int_0^\infty x^n e^{-x^{12}\sin^2x}dx$ is divergent for any natural numbers $n$? My attempt: Cauchy criteria near $2k\pi$ for integers $k$.
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$$\begin{align}\int_{k\pi}^{(k+1)\pi}e^{-x^{12}\sin^2x}\,\mathrm dx &= \int_0^\pi e^{-(k\pi+x)^{12}\sin^2x}\,\mathrm dx \\ &> \int_0^\pi e^{-((k+1)\pi)^{12}x^2}\,\mathrm dx \\ &= \int_0^\pi e^{-\pi^{10}((k+1)^6x)^2}\,\mathrm dx \\ &> \int_0^{\frac1{(k+1)^6}} e^{-\pi^{10}((k+1)^6x)^2}\,\mathrm dx \\ &> \int_0^{\frac1{(k+1)^6}} e^{-\pi^{10}}\,\mathrm dx \\ &=\frac{e^{-\pi^{10}}}{(k+1)^6} \end{align} $$ so that you will readily find that the integral diverges at least for $n\ge 6$.
Hagen von Eitzen
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For a more precise result, I think one can use Laplace's method to get $\int_{k \pi-\pi/2}^{k\pi+\pi/2} x^n \exp(-x^{12}\sin^2(x))\sim C (k \pi)^{n-6}$ when $k\to\infty$. – Gribouillis Jan 03 '21 at 14:50
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It seems that for $1\leq n<6$, the integral is also divergent. But I can not prove it. – xldd Jan 04 '21 at 00:15