calculus - Verify $limlimits_{nrightarrowinfty}int^infty_0 ! e^{-nx} sin(e^x) , mathrm{d}x = 0.$

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?

Answer

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 $n \to +\infty$.