generalized functions & operators I am dealing with a function $f(r) $that behaves like ~ $\frac{1}{r}$ when approaching zero. When I take the Laplacian of this guy and then integrate the result ([0,$\infty$]) I get some additional contribution from the delta function part of $\nabla^2 f$(r), which is not obvious at first glance.
The function $f(r)$ is not analytic at $r=0$ and this gives rise to a generalized function when the operator is acting on it. 
How do I explain it using rigorous mathematical language? Can anyone point out useful theorems?
 A: We can define the Laplacian of such functions $f$ as a signed measure (that is, a measure that can assign positive or negative values to sets). Precisely, the generalized/distributional Laplacian $\Delta f$ is a measure $\mu$ such that for any smooth compactly support function $\varphi$ we have $$\int f\, \Delta\varphi =\int \varphi\,d\mu \tag1$$ 
When $f$ is twice differentiable, $\mu$ is just a continuous measure with density $\Delta f$. In general, it may have singular part. In your example (assuming you work in $\mathbb R^3$) one can calculate $\Delta |x|^{-1}$ by applying Green's identity in a spherical shell $r<|x|<R$ and letting $r\to 0$. Choose $R$ large so that the support of $\varphi$ is within $|x|<R$. The normal derivative in direction away from the origin is denoted by $\frac{\partial  }{\partial n} $ below.
$$\begin{split}
\int_{r <|x|<R} |x|^{-1}\, \Delta\varphi(x) &= 
\int_{r <|x|<R} (|x|^{-1}\, \Delta\varphi(x) - \varphi(x)\Delta(|x|^{-1})) \\
&= \int_{|x|=r} \left( \frac{\partial |x|^{-1} }{\partial n} \varphi(x) - \frac{\partial \varphi }{\partial n} |x|^{-1}  \right) \\ 
& =-r^{-2}\int_{|x|=r} |x|^{-2}  \varphi(x) + r^{-1}\int_{|x|=r} \frac{\partial \varphi }{\partial n} \\  &\to -4\pi \varphi(0) +0 = \int \varphi\,d\mu
\end{split}
\tag2$$ 
where $\mu$ is $-4\pi\delta_0$.
A useful theorem: $\Delta f$ is a nonnegative measure if and only if $f$ is subharmonic. In this example $\Delta_f$ is nonpositive, hence $f$ is superharmonic.
Since the Laplacian is a linear operator, you can deal with your function  (which apparently has other things besides $1/r$ by writing it $f(x)=1/|x|+g(x)$ where $g$ may be sufficiently nice to have Laplacian in the classical sense.
