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Integrals and continuity

  1. Jul 30, 2007 #1

    daniel_i_l

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    Gold Member

    1. The problem statement, all variables and given/known data
    Prove that the function [tex]\int^{1}_{x}\frac{sin t}{t}dt[/tex] is uniformly continues in (0,1).


    2. Relevant equations



    3. The attempt at a solution

    First if all, I defined f(x) as sin(x)/x for x=/=0 and 0 for x=0. So f is continues in [0,1]. Now [tex]G(x) = \int^{x}_{1}f(t) dt[/tex] Is defined and continues in [0,1] so it's uniformly continues in (0,1). But in (0,1)
    [tex]G(x) = \int^{x}_{1}\frac{sin t}{t}dt[/tex] And since that's UC in (0,1), then also [tex]-\int^{x}_{1}\frac{sin t}{t}dt = \int^{1}_{x}\frac{sin t}{t}dt[/tex] is UC in (0,1). Is that enough?
    Because in the book they used the fact that G(x) has a derivative in [0,1] and that G'(x) = f(x). Then they used the fact that |f(x)|<=1 (meaning that the derivative of G is bounded) to prove that G(x) is UC.
    What's wrong with my way?
    Thanks.
     
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  3. Jul 30, 2007 #2

    Dick

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    I think if you want f(x)=sin(x)/x to be continuous, you'll want to define f(0)=1. Other than that, there is nothing wrong with proving it that way. If the integral is continuous on [0,1], then it is uniformly continuous on (0,1). I don't know why you are messing around with changing the limits of integration though.
     
  4. Jul 30, 2007 #3

    daniel_i_l

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    Gold Member

    Yeah, it's supposed to be f(0)=1. Thanks for your confirmation.
     
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