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$$ f(x) \ = \ \int_{x}^{x+1} \sin(e^t) \, dt. $$ Show that $$ e^x | f(x) | \ < \ 2 $$ and that $$ e^x f(x) = \cos(e^x) - e^{-1} \cos(e^{x+1}) + r(x), $$ where $|r(x)|< C e^{-x}$ for some constant $C$.

Using Integration by parts, I showed that $$ f(x) = \frac{\cos(e^x)}{e^x} - \frac{\cos(e^{x+1})}{e^{x+1}} - \int_{e^x}^{e^{x+1}} \frac{\cos u}{u^2} \, du . $$ Thus, in my calculation, $$ r(x) \ = \ -e^x \int_{e^x}^{e^{x+1}} \frac{\cos u}{u^2} \, du . $$ But I can't prove that $|r(x)| < C r^{-x}$ for some $C$.

Is it typo???? (We can show that $|r(x)| < 2 e^{-1}$.)

user108219
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1 Answers1

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let $e^x=u$,then \begin{align*} |f(x)|&=|\int_{x}^{x+1}\sin{(e^t)}dt|=|\int_{e^{x}}^{e^{x+1}}\dfrac{1}{u}d\cos{u}|\\ &=\left|\dfrac{\cos{e^{x+1}}}{e^{x+1}}-\dfrac{\cos{e^x}}{e^x}+\int_{e^{x}}^{e^{x+1}}\dfrac{\cos{u}}{u^2}du\right|\\ &\le|\dfrac{\cos{e^{x+1}}}{e^{x+1}}|+|\dfrac{\cos{e^x}}{e^x}|+\int_{e^x}^{e^{x+1}}\dfrac{1}{u^2}du\\ &=\dfrac{2}{e^x} \end{align*}

math110
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