Interpreting the logarithm as a sum of simple poles along the negative real axis

Question

I've heard it remarked that you can basically consider $\log z$ to be a function which has simple poles everywhere on the negative real axis (with a constant "residue density" at each pole). This would be something like $$ \log z = \int_0^\infty \frac{dx}{z + x} $$ But of course the integral on the right-hand side actually diverges. We actually get $$ \int_0^\infty \frac{dx}{z + x} = \lim_{b \to \infty} \int_0^b \frac{dx}{z + x} = \lim_{b \to \infty} \left( \log(z + b) - \log(z) \right) = \infty $$

In physics, there are a variety of methods for subtracting out the divergent part of such a limit to get a finite answer (various flavors of regularization and renormalization). I'm wondering whether there is a standard approach here so that something similar can be done to "rescue" the first equation above from the divergent part of the integral.

Another way of phrasing the problem above is that I showed that the Stieltjes transform of a constant on the interval $(-\infty, 0]$ does not exist. But perhaps there is another density function $\rho$ so that $\log z$ is the Stieltjes transform of $\rho$. $$ \log z = \int_0^\infty \frac{\rho(x)}{z + x} dx $$ What is $\rho$? Well, the Stieltjes inversion formula says that it should be given by $$ \begin{align} \rho(x) &= \lim_{\epsilon \to 0} \frac{\log(x+i\epsilon) - \log(x-i\epsilon)}{2\pi i} \\ &= \frac{(\log |x| + \pi i) - (\log |x| - \pi i)}{2\pi i} \\ &= 1 \end{align} $$

But this gets me exactly back to the integral that I started with, which is divergent! Hopefully I am just missing something obvious.

Edit: Alternate statement of question

There has been a lot of confusion in the comments below about what I am looking for, so let me restate it in a very narrow way. I would be satisfied with either of the following:

1. A sequence of meromorphic functions $f_n(z)$ with simple poles along the negative real axis with the following properties:

   a. The poles become dense in the limit $n \to \infty$.

   b. $\lim_{n \to \infty} f_n(z) = \log z$

2. A proof that there is no such sequence.

Answer 1

The correct formula is $$\log z = \lim_{T\to \infty} \log(z)-\log(1+\frac{z-1}{T+1})$$ $$=\lim_{T\to \infty} (\log(T+1)-\log(1))-(\log(T+z)-\log(z))$$ $$= \lim_{T\to \infty} \int_0^T (\frac1{1+x}-\frac{1}{z + x}) dx=\int_0^\infty (\frac1{1+x}-\frac{1}{z + x}) dx$$

There is no distribution such that $\log (x+iy) = \frac1{x+iy}\ast \rho$ because this distribution would have to be $C+1_{x>0}$ yielding a divergent integral.

Answer 2

By fiddling around with my original suggestion above, I have one solution that meets my criteria: $$ f_n(z) = - \sum_{k=0}^{n} \frac{1}{n\frac{z}{1-z} + k} $$ This has poles at $z=-\frac{k}{n-k}$ for $0 \le k < n$ which becomes dense on the negative real axis as $n \to \infty$.

If we let $a=\frac{nz}{1-z}$, then $$ f_n(z) = \psi(a) - \psi(a+n+1) $$ where $\psi$ is the digamma function. As $n \to \infty$, $a \to \infty$, and the leading asymptotic behavior of $\psi(z)$ is $\log z$ as $z \to \infty$. So we get $$ f_n(z) \sim \log\left( \frac{a}{a+n+1} \right) \sim \log\left( \frac{1}{1 + \frac{1-z}{z} + 0} \right) = \log z $$

This works in a technical sense, but it seems more complex than it needs to be. I would accept any answer which is more elegant, especially if it can be expressed in a Stieltjes-transform-like form.