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Improper integral convergence and implications of infinite limits

  1. Apr 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Let f be a continuous function on [1,∞) such that [itex]\lim_{x\rightarrow ∞}[/itex]f(x)=α. Show that if the integral [itex]\int^{∞}_{1} f(x)dx[/itex] converges, then α must be 0.

    2. Relevant equations
    Definition of an Improper Integral
    Let f be a continuous function on an interval [a,∞). then we define [itex]\int^{∞}_{a} f(x)dx[/itex] to be [itex]\lim_{b \rightarrow ∞}\int^{b}_{a} f(x)dx[/itex] if this limit exists. In this case, the integral is said to be convergent.

    3. The attempt at a solution

    My attempt at this solution was first to start with the epsilon-delta version of a limit of a function. That is, since [itex]\lim_{x\rightarrow ∞}[/itex]f(x)=α, then [itex]\forall \epsilon[/itex]>0 with [itex]x_{0}\in [/itex] [1,∞) [itex]\exists~ \delta[/itex]>0 s.t. |f(x) - α | whenever |x-[itex]x_{0}[/itex]|<[itex]\delta[/itex]. Then, I assumed that α was not equal to zero. From there I split up my inequality so that

    -ε<f(x) - α< ε [itex]\Rightarrow[/itex] α - ε<f(x)< ε + α [itex]\Rightarrow[/itex] [itex]\int^{∞}_{1}[/itex](α - ε)dx < [itex]\int^{∞}_{1} f(x)dx[/itex] < [itex]\int^{∞}_{1} (α + ε)dx[/itex]

    From here, I am getting stuck in my logic and I think that perhaps this is not the route I want to go. I would appreciate it if I could be offered any hints on where to go from here, or on whether or not I should start over. Thanks!
     
    Last edited by a moderator: Apr 9, 2012
  2. jcsd
  3. Apr 9, 2012 #2

    tiny-tim

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    Hi Szichedelic! :smile:
    No, for ∞, your δ is a number "near ∞" …

    we usually use N instead, and we look for an N such that x > N => |f(x) - α| < ε :wink:
     
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