I want to find limit of $\displaystyle\int_0^\infty ne^{-nx}\sin\left(\frac{1}{x}\right)\;dx$ as $n\to\infty$ if it exists or to prove that it doesn't exist. I see that $ne^{-nx}\sin\left(\frac{1}{x}\right)\to 0$ for all $x>0$ and that the convergence is uniform on $[a,\infty)$ for all $a>0$. That implies $\displaystyle\int_a^\infty ne^{-nx}\sin\left(\frac{1}{x}\right)\;dx\to 0$ as $n\to\infty$ for all $a>0$. Can anyone tell me what the next step is or if I'm on the wrong track? Thanks.
$\int_0^\infty ne^{-nx}\sin\left(\frac1{x}\right)\;dx\to ?$ as $n\to\infty$
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real-analysis
measure-theory
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1As you are tagging this "measure-theory", why don't you consider using dominated convergence theorem? Uniform convergence may not be a good idea since swapping integration/limit may not hold for unbounded interval, even under uniform convergence. – 2011-08-24
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1@user, if you recommend to use that there exists a so-called specific integer $m$ such that $ne^{-nx}\le me^{-mx}$ uniformly over $x$ and $n$ (or maybe every $n$ large enough), then you might reconsider: this is false. – 2011-08-24
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0@Didier: Oh yeah, you are right. Sorry about that. – 2011-08-25