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Thorycal Issue with Improper Integrals

  1. Feb 4, 2013 #1
    So, like i said in the Title this more of a thoerycal question. In my university notebook i have written that an integral to converge has to happen the next:

    1. The f has to be bounded (if not its just a dot)
    2.The interval has to be finit.

    [THIS IS WHAT IT'S WRITTEN IN MY NOTEBOOK]

    See, my really issue is what it means to be bounded. If has to be in an interval, or if has to have Upper and Lower bounds. And why does it say that the interval has to be finit if there are integral that are definite betwen 0 and infinity, for example and converge.

    ------------------------------------------------------------------

    For example:

    [itex]\int^{infinty}_{1} \frac{1}{t} \, dt [/itex]

    It's the function [itex]f(x) = \frac{1}{t}[/itex] or [itex]ln|t| + C[/itex] that has to be bounded.

    Thanks
     
    Last edited: Feb 4, 2013
  2. jcsd
  3. Feb 4, 2013 #2

    Mark44

    Staff: Mentor

    Doesn't your book have a definition of this term? It doesn't have anything to do with dots, as you said above.
    The above doesn't make sense. On the left side you have an improper definite integral. The right side is the antiderivative of 1/t. In other words
    $$\int \frac{dt}{t} = ln|t| + C$$


    What has to

    ##\int_1^{\infty} \frac{dt}{t} ## is not a function. What has to be bounded is the integrand, the function you are integrating.
     
  4. Feb 4, 2013 #3
    Sorry for my mistakes i confused, but thank you. But i ask you again, what really bothers me, if bounded means to be in an interval or ther have upper and lower bound. I know it sounds really stupid and maybe is obious but i dont get it. I wrote that integral beacuse the interval it's not finit, all the contrary, and i dont know if that function i'm integrating [itex]f(x) = \frac{1}{t}[/itex] its bounded, again because i dont know what it really means.

    Thanks and very sorry for my english.
     
  5. Feb 4, 2013 #4

    Mark44

    Staff: Mentor

    On the interval [1, ∞), f(t) = 1/t is bounded. In fact it is bounded above by 1, because for any t ≥ 1, 1/t ≤ 1. On the same interval, f(t) is bounded below by 0, since f(t) > 0 for all t ≥ 1.

    A function f is bounded above on an interval if there is some number M such that f(t) ≤ M for all t in that interval. The definition for bounded below is similar.
     
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