# Summability of $\frac{1}{\|\boldsymbol{r}-\boldsymbol{l}\|^2}$ on $V\times[a,b]$

Let $\boldsymbol{r}$ represent a point of $\mathbb{R}^3$ with two components fixed and one, say $r_k$, free to vary on $[a,b]$, and let $V\subset\mathbb{R}^3$ be a measurable, according to the tridimensional Lebesgue measure, and bounded subset.

I am trying to understand whether the Lebesgue integral $$\int_V\int_{[a,b]}\frac{1}{\|\boldsymbol{r}-\boldsymbol{l}\|^2}d\mu_{r_k}d\mu_{\boldsymbol{l}}$$ $$=\int_V\int_{[a,b]}\frac{1}{\sqrt{(r_i-l_i)^2+(r_j-l_j)^2+(r_k-l_k)^2}}d\mu_{r_k}d\mu_{\boldsymbol{l}}$$or equivalently$$\int_{[a,b]}\int_V\frac{1}{\|\boldsymbol{r}-\boldsymbol{l}\|^2}d\mu_{\boldsymbol{l}}d\mu_{r_k}\quad\text{ or }\quad\int_{[a,b]\times V}\frac{1}{\|\boldsymbol{r}-\boldsymbol{l}\|^2}d\mu_{\boldsymbol{l}}\otimes\mu_{r_k}$$exist finite. All the measures are intended as the usual $n$-dimensional Lebesgue measures. Does it exist and, if it does, how can it be proved?

I do see that, for all fixed $\boldsymbol{r}$, $\int_V\frac{1}{\|\boldsymbol{r}-\boldsymbol{l}\|^2}d\mu_{\boldsymbol{l}}$ exists finite by chosing a ball $B(\boldsymbol{r},R)\supset V$ and calculating $$\int_V\frac{1}{\|\boldsymbol{r}-\boldsymbol{l}\|^2}d\mu_{\boldsymbol{l}}\le\int_{B(\boldsymbol{r},R)}\frac{1}{\|\boldsymbol{r}-\boldsymbol{l}\|^2}d\mu_{\boldsymbol{l}}=\lim_{\varepsilon\to 0}\,\mathscr{R}\int_{B(\boldsymbol{r},R)\setminus B(\boldsymbol{r},\varepsilon)}\frac{1}{\|\boldsymbol{r}-\boldsymbol{l}\|^2}d^3l=4\pi R$$ where I use the notation $\mathscr{R}$ for the Riemann integral, which I have calculated by using spherical coordinates. But here, $r_k$ is free to vary in $[a,b]$, which prevents me from using this argument. I heartily thank any answerer.

• Choose $R$ large enough and you have an upper bound for $\int_V \frac{1}{\lVert r-l\rVert^2}\,d\mu_l$ that is independent of $r_k\in [a,b]$. So $$\int_{[a,b]}\int_V \frac{1}{\lVert r-l\rVert^2}\,d\mu_l\,d\mu_{r_k} < +\infty.$$ – Daniel Fischer Mar 5 '16 at 14:22
• What one needs (for this argument) is that there is an $\boldsymbol{x}$ and an $R$ with $\boldsymbol{r} - V = \{\boldsymbol{r} - \boldsymbol{l} : \boldsymbol{l} \in V\} \subset B(\boldsymbol{x},R)$ for all pertinent $\boldsymbol{r}$. That indeed follows from the boundedness of the segment. Note however that even $$\int_{-\infty}^{\infty} \int_V \frac{1}{\lVert \boldsymbol{r} - \boldsymbol{l}\rVert^2}\,d\mu_{\boldsymbol{l}}\,d\mu_{\boldsymbol{r}_k} < +\infty,$$ which cannot be obtained with only this argument. For that, use this argument for $\boldsymbol{r}$ "close to $V$", and note that – Daniel Fischer Mar 5 '16 at 15:09
• $$\int_V \frac{1}{\lVert \boldsymbol{r} - \boldsymbol{l}\rVert^2}\,d\mu_{\boldsymbol{l}} \sim \frac{\mu(V)}{\operatorname{dist}(\boldsymbol{r}, V)^2}$$ for $\boldsymbol{r}$ "far away from $V$". These arguments show that the integral is finite, but that doesn't help evaluating the integral. Except for some special cases, I don't think you can explicitly evaluate it, but numerical integration should work well enough if a value is needed. – Daniel Fischer Mar 5 '16 at 15:09
• Not that it matters, but I should have left out $\boldsymbol{x}$ and used $0$, that's more convenient. (Clearly, $B(\boldsymbol{x},R) \subset B(0,R + \lVert\boldsymbol{x}\rVert)$, so we can always pick $\boldsymbol{x} = 0$ if we want.) We have, by translation-invariance of the Lebesgue measure, and then positivity $$\int_V \frac{1}{\lVert\boldsymbol{r} - \boldsymbol{l}\rVert^2}\,d\mu_{\boldsymbol{l}} = \int_{\boldsymbol{r} - V} \frac{1}{\lVert\boldsymbol{k}\rVert^2}\,d\mu_{\boldsymbol{k}} \leqslant \int_{B(0,R)}\frac{1}{\lVert\boldsymbol{k}\rVert^2} \,d\mu_{\boldsymbol{k}} =: C,$$ – Daniel Fischer Mar 5 '16 at 20:42
• and that latter is a bound independent of $\boldsymbol{r}$. Now clearly $\int_{[a,b]} C\,d\mu_{\boldsymbol{r}_k} = C\dot (b-a) < \infty$, so we have the finiteness of the integral. – Daniel Fischer Mar 5 '16 at 20:42