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How to rigorously motivate the following formula

  1. Mar 9, 2015 #1
    I have encountered the following formula a couple of times (always in a physics context, of course..)
    [itex]\int_{0}^\infty \frac{dt}{t}e^{-tx}=-\log x[/itex]
    Formally one can "derive" this formula by noting that
    [itex]\log x=\int \frac{dx}{x}=\int dx \int_0^\infty dte^{-xt}=-\int_0^\infty \frac{dt}{t}e^{-xt}[/itex]
    But the t integral obviously diverges. So there must be some regularization of this integral but this is never explained (and sometimes they write that the integral is from ##0^+## instead of 0, whatever that means).
  2. jcsd
  3. Mar 9, 2015 #2


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    From a mathematical point of view, this doesn't look right. [itex]logx=\int_1^x \frac{du}{u}[/itex]. Therefore the final integral should be:
    [itex]-\int_0^{\infty} \frac {dt}{t}(e^{-xt}-e^{-t})[/itex], which is convergent.
  4. Mar 9, 2015 #3


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    Set ##x = 1## and you get:

    [itex]\int_{0}^\infty \frac{dt}{t}e^{-t}=0[/itex]

    Which can't be right.

    In any case, the LHS is positive and the RHS is negative for ##x > 1##.
  5. Mar 9, 2015 #4
    I agree with everything you are saying, but this is a formula I see quite often but I never understand what it means due to these issues. Just to throw in an example, look at page 5 before equation 1.11 of the following paper
  6. Mar 9, 2015 #5


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    I don't see the relation between the formula that uses a sum, an additional integral, some function K and other things but no exponential, and your function (which has been shown to be wrong).
  7. Mar 9, 2015 #6


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    Try offering it pizza. Always works for me :smile:
  8. Mar 9, 2015 #7
    The additional sum is the only difference. Just substitute the equation just before to get rid of the inner three dimensional integral.

    I am not making this up, I have seen this divergent integral popping up several times without any comments about the obvious fact that it is wrong...
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