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Integral inequality

  • #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?
 

Answers and Replies

  • #2
PeroK
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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?
Let ##f(x) = 1 - x + \int^x_1 \frac{\sin t}{t} \ dt##
 
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  • #3
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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?
 
  • #4
PeroK
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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?
What's stopping you?
 
  • #5
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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. ;(
 
  • #6
PeroK
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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. ;(
What's ##f'(x)##?
 
  • #7
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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} ##
 
  • #8
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And when ##x > 1##?
 
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  • #9
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Not the same?
 
  • #10
PeroK
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Not the same?
When ##x > 1## then ##f'(x) < 0##. You should be able to see that sort of thing.
 
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  • #11
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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
 
  • #12
PeroK
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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 is always smaller than 1?

EDIT: see edit
Yes, ##f(1) = 0##, so if ##f'(x) < 0## for ##x > 1## you are done.
 
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  • #13
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ok, thanks!
 

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