# What is meant by the following divergent formula?

I have encountered the following formula a couple of times (in different physics contexts which I do not have a good understanding of)

$$\int_{0}^\infty \frac{dt}{t}e^{-tx}=-\log x$$

Formally one can "derive" this formula by noting that $$\log x=\int \frac{dx}{x}=\int dx \int_0^\infty dte^{-xt}=-\int_0^\infty \frac{dt}{t}e^{-xt}$$ However, if one does this carefully I think one should obtain something like $$\int_{0}^\infty \frac{dt}{t}(e^{-tx}-e^{-t})=-\log x$$ which is convergent, but this is not how it is ever stated! To give an example, look at page 5 before equation 1.11 of the following paper http://arxiv.org/pdf/0804.1773v1.pdf (just substitute the equation just before to obtain the above formula). How should one use and interpret this equation? Is there some hidden regularization scheme which is obvious to everyone except me?

• is that $log x$ or rather $ln x$ ? – docscience Mar 10 '15 at 23:55
• This is one of those cases where physicists happily interchange limits when it in fact can't be done. However, without understanding the context (and I don't understand it), I can't say what the justification is. – Javier Mar 11 '15 at 2:33

My guess would be that almost always in physics $x$ stands for a dimensionful quantity, then the logarithm cannot be well defined unless for a dimensionless ratio, i.e. log $(x/x_0)$. In which case you can apply the given formula which becomes perfectly convergent $$-\text{log }\left(\frac{x}{x_o}\right) =\int_0^{\infty}\frac{dt}{t}\left(e^{-tx} - e^{-tx_o} \right)$$