How to rigorously motivate the following formula

[Kurret]

I have encountered the following formula a couple of times (always in a physics context, of course..)
\int_{0}^\infty \frac{dt}{t}e^{-tx}=-\log x
Formally one can "derive" this formula by noting that
\log x=\int \frac{dx}{x}=\int dx \int_0^\infty dte^{-xt}=-\int_0^\infty \frac{dt}{t}e^{-xt}
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).

 

[mathman]

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

 

[PeroK]

I have encountered the following formula a couple of times (always in a physics context, of course..)
\int_{0}^\infty \frac{dt}{t}e^{-tx}=-\log x

Set $x = 1$ and you get:

\int_{0}^\infty \frac{dt}{t}e^{-t}=0

Which can't be right.

In any case, the LHS is positive and the RHS is negative for $x > 1$.

 

[Kurret]

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

Set $x = 1$ and you get:

\int_{0}^\infty \frac{dt}{t}e^{-t}=0

Which can't be right.

In any case, the LHS is positive and the RHS is negative for $x > 1$.

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
http://arxiv.org/pdf/0804.1773v1.pdf

 

[mfb]

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).

 

[phinds]

How to rigorously motivate the following formula

Try offering it pizza. Always works for me

 

[Kurret]

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).

The additional sum is the only difference. Just substitute the equation just before to get rid of the inner three dimensional integral.

and your function (which has been shown to be wrong).

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...