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Proving function is improper riemann integrable

  1. Jan 5, 2007 #1
    1. The problem statement, all variables and given/known data

    let f:[0,oo) -> R be given by f(x) = sin(x) / x for x>0 and f(0) = c. Prove that f is improper riemann integrable without computing the integral explicitly

    3. The attempt at a solution

    I've attempted to find a upperbound for f(x) such that the integral does not diverge. The most simple one is to use the fact that sin(x) <= 1 for all x, but this gives a divergent integral.

    I've already proved that f is Lebesgue measurable for every c in R. So I could turn the integral [tex]\int_0^R f(x) dx[/tex] into a Lebesgue integral and then use one of the convergence theorems to try to show with them that the integral does not diverge. But I haven't succeded in doing this :(

    Can anyone help me out?
  2. jcsd
  3. Jan 5, 2007 #2


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    As you've probably noticed, the integral of the absolute value of your function diverges. The fact an improper integral can be defined means you must be able to show that the positive parts can be combined with the negative parts to get a partial cancellation that can converge. Draw a graph and think about estimating. Does that help?
  4. Jan 5, 2007 #3


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    Hoi Piet.

    Use the Cauchy criterion!:
    The improper integral [tex]\int_a^\infty f(x)dx[/tex] converges if and only if [itex]\forall \epsilon>0 \exists b>a[/itex] such that
    [tex] c,d > b \Rightarrow \left|\int_c^d f(x)dx \right| < \epsilon[/tex]

    You can use integration by parts to show that:
    [tex]\left| \int_c^d \frac{\sin x}{x}dx\right|\leq 2\left(\frac{1}{c}+\frac{1}{d}\right)[/tex]
    Last edited: Jan 5, 2007
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