Exercise 6.14

Question

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Exercise 14: Deal similarly with $$f(x)=\int_x^{x+1}\sin e^t\,dt.$$ Show that $$e^x\big|f(x)\big|<2$$ and that $$e^xf(x)=\cos e^x-e^{-1}\cos e^{x+1}+r(x),$$ where $\big|r(x)\big|<Ce^{-x}$,
    for some constant $C$.

analambanomenos · Accepted Answer

Substituting $u$ for $e^t$ and integrating by parts, we get
    \begin{align*}
        f(x) &= \int_x^{x+1}\sin e^t\,dt \
        &= \int_{e^x}^{e^{x+1}}\frac{\sin u}{u}\,du \
        &= \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
    \end{align*}
    Hence, substituting $-1$ for $\cos u$ in the integral, we get 
    \begin{align*}
        e^xf(x) &< \cos e^x-\frac{\cos e^{x+1}}{e}-e^x\int_{e^x}^{e^{x+1}}u^{-2}\,du \
        &= (1+\cos e^x) - e^{-1}(1+\cos e^{x+1}),
    \end{align*}
    so that $e^x\big|f(x)\big|<2(1+1/e)$, which isn't quite 2 (more work would need to be done to get this below 2). Doing another integration by parts, we get
    \begin{align*}
        r(x) &= -e^x\int_{e^x}^{e^{x+1}}\frac{\cos u}{u^2}\,du \
        &= e^x\Biggl(\frac{\sin e^x}{e^{2x}}-\frac{\sin e^{x+1}}{e^{2x+2}}-2\int_{e^x}^{e^{x+1}}\frac{\sin u}{u^3}\,du\Biggr) \
        &< \frac{\sin e^x}{e^x}-\frac{\sin e^{x+1}}{e^{x+2}}+2e^x\int_{e^x}^{e^{x+1}}u^{-3}\,du \
        &= \frac{1+\sin e^x}{e^x}-\frac{1+\sin e^{x+1}}{e^{x+2}},
    \end{align*}
    so that $\big|r(x)\big|<2(1+1/e^2)e^{-x}$.

Exercise 6.14

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Exercise 6.14

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