How to derive a logarithmic potential from Newtonian? Suppose we believe that the formula for Newtonian potential in $R^3$ is correct:
$\varphi(\bar{x}) = \frac{1}{|x|} = \frac{1}{r}$, disregarding the constant. 
What is the justification of the fact that on $R^2$ this potential reduces to $\varphi(\bar{x}) = -\log( |x| ) = -\log( r )$?
Essentially, I could imagine two reductions: a potential of a point-mass or a potential of a string-mass.
The point one:
$$\int_{x \in R}\delta(x-x_0)\frac{1}{\sqrt{x^2+y^2+z^2}} = \frac{1}{\sqrt{{x_0}^2+y^2+z^2}} $$ or if we want to renormalize it, say, $$\frac{1}{\sqrt{{x_0}^2+y^2+z^2}} - \varphi(x_0) = \frac{1}{\sqrt{{\frac{1}{\phi(x_0)}}^2+y^2+z^2}} - \varphi(x_0)$$
This is a projection of a Newtonian potential on a 2-D plane $(x_0, y, z)$. Doesn't look like a logarithm...
If we look at the string, lying on the X axis:
$$\delta(y,z)*\frac{1}{\sqrt{x^2+y^2+z^2}} = \int_{-\infty}^{\infty}\frac{dx}{\sqrt{{x}^2+y^2+z^2}}$$ 
I don't really understand how to compute the last expression. Something like: (I am not sure about this formula.)
$$\int_{-\infty}^{\infty}\frac{dx}{\sqrt{{x}^2+y^2+z^2}} = v.p \int\frac{dx}{\sqrt{{x}^2+y^2+z^2}} + \int\frac{\delta(x)dx}{\sqrt{{x}^2+y^2+z^2}}$$ 
The right summant is no logarithmic, and the left doesn't even converge!
However, just blindly plugging it into the table of integrals would give me something like 
$ \log| x + \sqrt{ x^2 + (r=y^2+z^2)^2}|/_{-\infty}^{\infty}$ 
Which is somehow related to the logarithm, but since it has an essential singularity at 0, it is not correct to use the Newton-Leibniz formula.
Could someone enlighten me or point to a right direction (m.b. book)?
 A: Your second attempt, the string-mass, is on the right track. If you started in $\mathbb{R}^n$ with $n\ge 4$, it would work out in a straightforward way: at distance $r$ from an infinite string, the potential is a constant multiple of
$$\int_{-\infty}^\infty \frac{1}{(r^2+z^2)^{(n-2)/2}}\,dz
=\int_{-\infty}^\infty \frac{1}{(1+u^2)^{(n-2)/2}}\,\frac{du}{r^{n-3}}
= C/r^{n-3}$$ using the substitution $u=z/r$.
The transition from $\mathbb{R}^3$ to $\mathbb{R}^2$ is more difficult because the potential theory has a different flavor in $\mathbb{R}^2$ (the potential is unbounded). As you noted, the  potential of a uniformly charged infinite string is infinite: the integral given above   diverges when $n=3$. In such a situation, one often uses regularization: subtract off the largest asymptotic term and extract information from the rest.   
Consider the finite string from $(0,0,-L)$ to $(0,0,L)$. Its   potential   at the point $(r,0,0)$  is equal to 
$$
\int_{-L}^L \frac{1}{\sqrt{r^2+z^2}}\,dz = \log\frac{\sqrt{r^2+L^2}+L}{\sqrt{r^2+L^2}-L}
= 2 \log(\sqrt{r^2+L^2}+L) - 2\log r$$ 
This expression tends to infinity at the rate $2\log L$. The latter quantity doesn't depend on $r$, so we may as well subtract it. (Recall that potential is defined up to an arbitrary constant; this subtraction is just a way of normalizing it). After subtracting $2\log L$  we get
$$2 \log\left(\sqrt{(r/L)^2+1}+1\right) - 2\log r\to 2\log 2-2\log r$$ 
i.e., the logarithmic potential.
