Test for integrability

  • Thread starter Palindrom
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Main Question or Discussion Point

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?
 
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Answers and Replies

StatusX
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No, take f(x)=1/(x ln(x)).
 
matt grime
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1/x isn't even a function from [0,infinity) to itself, never mind a measurable one.
 
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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...
 
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Ok, why is 1/(xln(x)) not integrable?

Another stupid question, I answered myself... Thanks!
 
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Gib Z
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What baffles me is that you define 1/0 to be 81...
 

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