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How to verify $\lim\limits_{n\rightarrow\infty}\int^\infty_0 \! e^{-nx} \sin(e^x) \, \mathrm{d}x = 0$?

My idea is to use the dominant convergence theorem with $f_n(x):= e^{-nx} \sin(e^x)$ and $f(x):=\lim\limits_{n\rightarrow\infty}f_n(x)$.

$\Rightarrow \lim\limits_{n\rightarrow\infty}\int^\infty_0 \! e^{-nx} \sin(e^x) \, \mathrm{d}x = \int^\infty_0 \! \lim\limits_{n\rightarrow\infty} e^{-nx} \sin(e^x) \, \mathrm{d}x = \int^\infty_0 \!\lim\limits_{n\rightarrow\infty}0\, \mathrm{d}x = 0$

Can I use this here?

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  • $\begingroup$ Note that dominated convergence also works fine. $\endgroup$ – Gabriel Romon Jun 18 '15 at 18:32
  • $\begingroup$ this could be of some interest here: en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma $\endgroup$ – Math-fun Jun 18 '15 at 18:39
  • $\begingroup$ The $sin(e^x)$ term may look intimidating, but actually it doesn't do anything ! You could replace it by any other function that is sufficiently well-behaved for small values of $x$, and get the same result. $\endgroup$ – M. Wind Jun 19 '15 at 0:15
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How to verify $\lim\limits_{n\rightarrow\infty}\int^\infty_0 \! e^{-nx} \sin(e^x) \, \mathrm{d}x = 0$

You may just observe that $$ \left|\int^\infty_0 \! e^{-nx} \sin(e^x) \, \mathrm{d}x\right|\leq \int^\infty_0 \! \left|e^{-nx} \right|\, \mathrm{d}x=\frac1n, \qquad (n>0), $$ and then let $n \to +\infty$.

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  • $\begingroup$ This helped, thank you. I was already sceptical about my ansatz, because I was lacking a dominant function... $\endgroup$ – kaos Jun 18 '15 at 16:28
  • $\begingroup$ @kaos The dominated function could have been $g(x)=e^{-x}|\sin(e^x)|$... since $|f_n(x)|\leq g(x)$, ( $n\geq 1$). But it seems easier to use the above answer. Thanks! $\endgroup$ – Olivier Oloa Jun 18 '15 at 16:30
  • $\begingroup$ facepalm I did not think about that one. Thank you. $\endgroup$ – kaos Jun 18 '15 at 16:33
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    $\begingroup$ You can use $e^{-x}$ to dominate the function and then integral of limits is the limits of the integral. $\endgroup$ – OKPALA MMADUABUCHI Jun 18 '15 at 16:47

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