An improper integral and its convegence I have an integral
$$I(\gamma)=\int\int d^3 \mathbf{r} \, d^3 \mathbf{r}' \frac{1}{|\mathbf{r}-\mathbf{r}'|+\gamma}$$
were $\gamma$ is a positive number, $\mathbf{r},\mathbf{r}' \in \mathbb{R}^3$, $\mathbb{R}^3$ is 3D Euclidean space 
Can the integral be solved analytically as a function of $\gamma$ and does it converge when $\gamma \rightarrow 0$
 A: The function $\phi(\vec r)$, as defined by the integral 
$$\phi(\vec r)=\int_{|\vec r'|\le R}\frac{1}{|\vec r-\vec r'|}d^3\vec r'$$
is the solution to Poisson's equation 
$$\nabla ^2\phi(\vec r)=
\begin{cases}
-4\pi\,\, \text{for}\,\,r\le R\\\\
0\,\,\,\,\,\,\,\,\,\, \text{for}\,\,r\ge R
\end{cases}
$$
with condition 
$$\lim_{|\vec r|\to \infty}\phi =0$$
The solution to the PDE is 
$$\phi(\vec r)=
\begin{cases}
\frac{4\pi}{3}\frac{3R^2-|\vec r|^2}{2}\,\, \text{for}\,\,r\le R\\\\
\frac{4\pi}{3}\frac{R^3}{|\vec r|}\,\,\,\,\,\,\,\,\,\, \text{for}\,\,r\ge R
\end{cases}
$$
for which one can see that 
$$\int_{\mathscr{R}^3}\phi(\vec r)d^3\vec r$$
does not converge.

NOTE:
For $\gamma \ne 0$, we can show that the integral of interest also diverges.  To that end, we let $R=|\vec r-\vec r'|$ and write
$$\begin{align}
\frac{1}{|\vec r-\vec r'|+\gamma}&=\frac{1}{R+\gamma}=\frac{1}{R}\left(\frac{1}{1+\gamma/R}\right)\\\\
&=\frac{1}{R}\left(1-\frac{\gamma}{R}+O\left(\frac{\gamma^2}{R^2}\right)\right)
\end{align}$$
from which previous analysis showed that the integral over $\vec r$ and $\vec r'$ of first term $1/R$ diverges.
A: There's no hope: The inner integral equals $\infty$ for each fixed $r.$ In other notation,
$$\int_{\mathbb {R}^3} \frac{1}{|x-a|+\gamma}\,dV(x) = \infty$$
for each fixed $a\in \mathbb {R}^3$ and $\gamma > 0.$ Here $x=(x_1,x_2,x_3)\in \mathbb {R}^3,$ and $dV$ is volume measure. Note the integral is independent of $a$ because volume measure is translation invariant. (That already shows your double integral equals $\infty.$) So just take $a=0.$ Then
$$\int_{|x|>1} \frac{dx}{|x|+ \gamma} = \sum_{n=1}^{\infty} \int_{n<|x|<n+1}\frac{dx}{|x|+ \gamma} \ge \sum_{n=1}^{\infty} \frac{1}{n+1+ \gamma}V(\{n<|x|<n+1\}).$$
Since $V(\{n<|x|<n+1\})$ is a constant times $(n+1)^3-n^3,$ we see the above series equals $\infty.$
