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Math Help - Intergral

  1. #1
    Junior Member tedii's Avatar
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    Intergral

    Let
     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)-\frac{cos(e^{x+1})}{e}+r(x)

    where  |r(x)|<Ce^{-x}, for some constant C.


    I got everything except I don't know where to get the r(x) part at all...
    Last edited by mr fantastic; December 10th 2009 at 06:07 PM. Reason: Fixed the latex: {} is used to delimit indices
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  2. #2
    Senior Member Shanks's Avatar
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    by letting u=e^t, we have:
    f(x)= \int_{x}^{x+1} sin(e^t) \, dt = \int_{e^x}^{e^{x+1}} \frac{\sin u}{u} \, du = e^{-y}\int_{e^x}^{e^{x+1}} \sin u \, du
    Thus
    |f(x)|\leq 2e^{-y}\leq 2e^{-x}
    where y belongs to the interval (x, x+1).
    From the above computation,I think, it should be bounded by a constant C, not Ce^{-x}. are you sure it is Ce^{-x}? I guess This boundry is too strong .
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  3. #3
    Junior Member tedii's Avatar
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    It does say that in Rudin, if you have it chapter 6 problem 14. But what you did helped. I went about proving the the <2 part a bit incorrectly. Thank you for your help though.
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  4. #4
    Senior Member Shanks's Avatar
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    I saw the probelm, maybe you are right, we need a tighter and better boundry.
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  5. #5
    Senior Member Shanks's Avatar
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    by using integration by parts twice, you will get the result.
    Last edited by Shanks; December 12th 2009 at 03:38 AM.
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