# 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 $\gamma \rightarrow 0$

• What is $dr$ for $r\in \mathbb{R}^3$? Jul 3, 2015 at 15:29
• What is $r\to-\infty$ or $r\to\infty$ for $r\in \Bbb R^3$? (I refer to the bounds of the integrals.) Jul 3, 2015 at 15:31
• I dont know, I guess $-\infty$
– Mike
Jul 3, 2015 at 15:33
• My point is that $-\infty$ and $\infty$ are not in $\Bbb R^3$, but are points in the extended real number system. Jul 3, 2015 at 15:34
• ok, lets consider whole Euclidean space only
– Mike
Jul 3, 2015 at 15:35

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.

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 $\infty.$