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Show that:

$$\int_0^\infty \frac {x^n e^{-x^2}}{1+x^2} $$

Converges for every value of $n$ ($n$ is a natural number).

I know how to show that the integrand goes to $0$ as $x$ goes to $\infty$, but it's not enough. I tried finding other integrals that converges that are bigger than the one in the question but failed. I thought also on using taylor expantion of $e^{-x^2}$ but failed also.

Can I get any help?

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Assuming that $n>-1$, you can write

$$0\le \int_0^\infty \frac{x^ne^{-x^2}}{1+x^2}dx\le \int_0^\infty x^ne^{-x^2}dx.$$ The latter integral converges and can be found explcitly (a change of variables $y=x^2$ is helpful).

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  • $\begingroup$ n is a natural number, edited. $\endgroup$ – Noam Mansur Jan 20 '16 at 17:19
  • $\begingroup$ @NoamMansur the answer still holds. $\endgroup$ – TZakrevskiy Jan 20 '16 at 17:20
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There's no problem in 0 so you should check it converges in $\infty$. Use the fact that if $f$ is a positive continuous function on $\mathbb{R}$ and $\exists\alpha >1,\,\lim\limits_{x\to\infty}x^\alpha f(x)=0$ then $\int_0^\infty f(x)dx$ converges.

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