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

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.

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Xander Henderson, Isaac Browne, GNUSupporter 8964民主女神 地下教會, Chris Custer
If this question can be reworded to fit the rules in the help center, please edit the question.

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 $\displaystyle\int_{M}^{\infty}e^{-x^{2}}x^{m}dx\leq\int_{M}^{\infty}\dfrac{1}{x^{m+2}}dx<\infty$.
• Unless $a=0$... – mathworker21 Apr 9 '18 at 18:35