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Integral inequality

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  1. Jan 5, 2017 #1

    Rectifier

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    The problem
    Show that ## 1-x+ \int^x_1 \frac{\sin t}{t} \ dt < 0## for ## x > 1 ##

    The attempt
    I rewrite the integral as
    ##\int^x_1 \frac{\sin t}{t} \ dt < x-1 ##

    This is about where I get. Can someone give any suggestions on how to continue from here?
     
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  3. Jan 5, 2017 #2

    PeroK

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    Let ##f(x) = 1 - x + \int^x_1 \frac{\sin t}{t} \ dt##
     
  4. Jan 5, 2017 #3

    Rectifier

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    Should I find the derivative to the function and plot it to see if the graph to the function is below the x-axis?
     
  5. Jan 5, 2017 #4

    PeroK

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    What's stopping you?
     
  6. Jan 5, 2017 #5

    Rectifier

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    Well, I tried but I stumbled upon this:
    ##1=\frac{\sin x}{x}## when I derived it and wanted to see where the derivative was 0. I don't think that I can solve that. ;(
     
  7. Jan 5, 2017 #6

    PeroK

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    What's ##f'(x)##?
     
  8. Jan 5, 2017 #7

    Rectifier

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    It's ##f'(x) = D \left(1-x+ \int^x_1 \frac{\sin t}{t} \ dt \right) = -1+ \frac{\sin x}{x} ##
     
  9. Jan 5, 2017 #8

    PeroK

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    And when ##x > 1##?
     
  10. Jan 5, 2017 #9

    Rectifier

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    Not the same?
     
  11. Jan 5, 2017 #10

    PeroK

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    When ##x > 1## then ##f'(x) < 0##. You should be able to see that sort of thing.
     
  12. Jan 5, 2017 #11

    Rectifier

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    How can I see that? Anything that I should look for? Does it have to do with the fact that the fraction ## \frac{\sin x}{x} ##is always smaller than 1 since ##\lim_{x \rightarrow \infty} \frac{\sin x}{x} = 0##?

    EDIT: see edit
     
  13. Jan 5, 2017 #12

    PeroK

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    Yes, ##f(1) = 0##, so if ##f'(x) < 0## for ##x > 1## you are done.
     
  14. Jan 5, 2017 #13

    Rectifier

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    ok, thanks!
     
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