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Show that $$I=\int_0^{\infty}e^{-a^2x^2}x^m\sin{nx}\,dx$$ is absolutely converges for $m>0$.

$\int_0^{\infty}|e^{-a^2x^2}x^m\sin{nx}|\,dx\leq e^{-a^2x^2}x^m$

I want to use comparison test. How to check the convergence of $\int_0^\infty e^{-a^2x^2}x^m$.

Please help me to solve the remaining. Also suggest me whether the problem can be solved by Dirichlet's Test.

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closed as off-topic by user21820, Xander Henderson, Isaac Browne, GNUSupporter 8964民主女神 地下教會, Chris Custer May 1 '18 at 15:50

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For $a\ne 0$, the $a$ can be set as $1$, or else, change of variable goes through.

For $e^{u}>u^{m+1}$ for large $u>0$, then $e^{-x^{2}}<x^{-2m-2}$ and we have $\displaystyle\int_{M}^{\infty}e^{-x^{2}}x^{m}dx\leq\int_{M}^{\infty}\dfrac{1}{x^{m+2}}dx<\infty$.

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  • $\begingroup$ Unless $a=0$... $\endgroup$ – mathworker21 Apr 9 '18 at 18:35
  • $\begingroup$ It seems that OP has missed to state that. $\endgroup$ – user284331 Apr 9 '18 at 18:39

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