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Ok so I have to evaluate the following: $$\lim_{n \to \infty} \int_0^{\infty} n^2\sin\left( \left(\frac{x}{n} \right)^2 \right) e^{-x^2} \,dx$$

Which is no problem at all when we put the limit in.

My problem is justifying the limit step. I am quite sure that I have to use Dominated Convergence Theorem- need to bound $|f_k|$'s by some $g$ that is integrable but I am having major issues figuring it out. Any help would be appreciated.

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Hint: $|\sin x | \leqslant |x| \implies n^2 |\sin[(x/n)^2]|e^{-x^2} \leqslant x^2e^{-x^2}$

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  • $\begingroup$ Yes but $x^2$ is not integrable. I could use it for (0,1) interval. But what about $(0, \infty ) $ ? So the limit actually bounds it? It is an increasing function? $\endgroup$ – John Smith Nov 24 '15 at 3:53
  • $\begingroup$ How about $x^2e^{-x^2}$? We have $|f_n(x)|$ dominated by the integrable $g(x) = x^2e^{-x^2}$. $\endgroup$ – RRL Nov 24 '15 at 3:54
  • $\begingroup$ If it is true for all n then the limit itself will do the job but I can't see how that second inequality holds. In particular why $n^2| sin[ (x/n)^2] | \leq x^2$ ? I guess it is obvious but I am missing the point somehow at 4 am.. $\endgroup$ – John Smith Nov 24 '15 at 4:00
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    $\begingroup$ The inequality $|\sin y| \leqslant |y|$ is well-known and holds for all $y$. Choose $ y = (x/n)^2$. Hence $n^2|\sin[(x/n)^2]| \leqslant n^2 (x/n)^2 = x^2$. $\endgroup$ – RRL Nov 24 '15 at 4:04
  • $\begingroup$ Shoot. Ok its all done. Thank you very much $\endgroup$ – John Smith Nov 24 '15 at 4:05

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