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Test for integrability

  1. Dec 30, 2006 #1
    It started out as an attempt to solve a HW question (which I also posted in the appropriate forum), but now I'm just curious as to the general case;

    Assume f>0 is a measurable function from [0,infinity) to itself. Then if xf(x) tends to zero as x tends to zero, there is a positive [tex]\epsilon[/tex] for which the integral of f over [tex][0,\epsilon ][/tex] is finite.

    This is following the intuition that while 1/x isn't integrable, multypling it by anything that tends to zero is.

    What do you say? True, not true?
     
    Last edited: Dec 30, 2006
  2. jcsd
  3. Dec 30, 2006 #2

    StatusX

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    No, take f(x)=1/(x ln(x)).
     
  4. Dec 30, 2006 #3

    matt grime

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    1/x isn't even a function from [0,infinity) to itself, never mind a measurable one.
     
  5. Dec 30, 2006 #4
    Ok, one by one:

    StatusX, 1/ln(x) doesn't tend to zero when x tends to zero, it tends to infinity... Edit: wait, it does, I'm an idiot.

    matt - It's almost everywhere defined, what's the problem? Define 1/0=81...
     
    Last edited: Dec 30, 2006
  6. Dec 30, 2006 #5
    Ok, why is 1/(xln(x)) not integrable?

    Another stupid question, I answered myself... Thanks!
     
    Last edited: Dec 30, 2006
  7. Dec 30, 2006 #6

    Gib Z

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    What baffles me is that you define 1/0 to be 81...
     
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