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Show limit of improper integral is 0

  1. Jan 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Suppose [itex]f[/itex] is real-valued, bounded, continuous, and non-negative and suppose [itex]\int_x^\infty f(t)\,dt[/itex] is convergent (is finite) for all [itex]x[/itex]. Is it true that
    [tex]\lim_{x\rightarrow \infty} {\int_x^\infty f(t)\,dt} = 0 \ ?[/tex]
    2. Relevant equations

    3. The attempt at a solution
    I can't think of a counterexample and it seems true intuitively, so I'm trying to prove it's true.
    Given [itex]\epsilon >0[/itex], I want to show there is [itex]M[/itex] such that
    [tex]\int_M^\infty f(t)\,dt < \epsilon \ .[/tex]
    I think [itex]\lim_{t\rightarrow \infty} {f(t)} = 0[/itex] although I am not sure how to prove this. Specifically, how would you prove that the improper integral would not exist if this limit did not exist?
    I'm not sure what to do at this point.
  2. jcsd
  3. Jan 6, 2012 #2


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    Science Advisor
    Homework Helper

    Try looking at the dominated convergence theorem.
  4. Jan 6, 2012 #3
    I don't know how that would help since the set we're integrating over is changing with the limit.
  5. Jan 6, 2012 #4
    Ah, so you can use that theorem??

    Note that



    [tex]I_{[x,+\infty[}:\mathbb{R}\rightarrow \mathbb{R}:t\rightarrow \left\{\begin{array}{cc} 0 & \text{if}~t<x\\ 1 & \text{if}~t\geq x \end{array}\right.[/tex]

    So what does the dominated convergence theorem tell you?
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