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Does this improper integral converge?

  1. Feb 5, 2005 #1
    Does this improper integral converge???

    I have an interesting problem that has been bothering me.

    Given:
    f(x)=1/x
    g(x)= (any continuous, non-negative function)
    g(x)<f(x)
    A = (a positive constant)

    I want to know:
    Does the integral of g(x) from A to +infinity converge? Or is it impossible to determine?
     
  2. jcsd
  3. Feb 5, 2005 #2

    dextercioby

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    Take for example [itex] g(x)=\frac{1}{x^{2}} [/itex] and see what happens...

    Daniel.
     
  4. Feb 5, 2005 #3
    g(x) does NOT need to be in form 1/(x^P)

    I already know that for any function of the form 1/(x^p), where p>1, it will converge. Thus, I know 1/(x^2) converges.

    HOWEVER, g(x) does NOT need to be in the form of 1/(x^p). g(x) can be ANY non-negative function such that g(x)<f(x). That's a completely different picture.
     
  5. Feb 6, 2005 #4
    Assumingly, you're asking whether or not a function like 1/(x+1) converges or something? With the criteria you mentioned, the function where f(x) > g(x) and a non-negative function could be either/or convergent/divergent.
     
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