# How to rigorously motivate the following formula

1. Mar 9, 2015

### 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).

2. Mar 9, 2015

### 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.

3. Mar 9, 2015

### PeroK

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 $x > 1$.

4. Mar 9, 2015

### Kurret

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

5. Mar 9, 2015

### Staff: Mentor

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).

6. Mar 9, 2015

### phinds

Try offering it pizza. Always works for me

7. Mar 9, 2015

### Kurret

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...