Does the following limit exists? $$
\lim_{a\to 0}\left\{a\int_{-R}^{R}\int_{-R}^{R}
\frac{\mathrm{d}x\,\mathrm{d}y}
{\,\sqrt{\,\left[\left(x + a\right)^{2} + y^{2}\right]
\left[\left(x - a\right)^{2} + y^{2}\right]\,}\,}\right\}
\quad\mbox{where}\ R\ \mbox{is a}\ positive\ \mbox{number.}
$$
 The integral exists, since you can take small disks around singularities and make a change of variables. It seems that the limit should be $0$, as $\lim_{a \to 0}\left[a\log\left(a\right)\right] = 0$.
 A: This is not an answer yet but just a few manipulations that might hopefully simplify things a bit. I might have made some calculation errors.
$$\int\limits_{-R}^{R}\int\limits_{-R}^{R}\frac{1}{\sqrt{\left[(x+a)^2+y^2\right]\left[(x-a)^2+y^2\right]}}\,dx\,dy \leq$$
$$\int\limits_{0}^{\sqrt{2}R}\int\limits_{0}^{2\pi}\frac{1}{\sqrt{\left(r^2+2ar\cos\theta+a^2\right)\left(r^2-2ar\cos\theta+a^2\right)}}r\,d\theta\,dr=$$
$$\int\limits_{0}^{\sqrt{2}R}\int\limits_{0}^{2\pi}\frac{1}{\sqrt{r^4-2a^2r^2\cos2\theta+a^4}}r\,d\theta\,dr=$$
$$\frac{1}{2}\int\limits_{0}^{2\pi}\int\limits_{0}^{2R^2}\frac{1}{\sqrt{\rho^2-2a^2\rho\cos2\theta+a^4}}\,d\rho\,d\theta=\frac{1}{2}\int\limits_{0}^{2\pi}\int\limits_{0}^{2R^2}\frac{1}{\sqrt{\left(\rho-a^2\cos2\theta\right)^2+a^4\sin^22\theta}}\,d\rho\,d\theta=$$
$$\frac{1}{2}\int\limits_{0}^{2\pi}\int\limits_{-a^2\cos2\theta}^{2R^2-a^2\cos2\theta}\frac{1}{\sqrt{\rho^2+a^4\sin^22\theta}}\,d\rho\,d\theta=$$
$$\frac{1}{2}\int\limits_{0}^{2\pi}\log\left(\frac{\sqrt{4R^4 - 4R^2a^2\cos2\theta+a^4} + 2R^2-a^2\cos2\theta}{2a^2\sin^2\theta}\right)\,d\theta \leq$$
$$\frac{1}{2}\int\limits_{0}^{2\pi}\log\left(\frac{4R^2+2a^2\sin^2\theta}{2a^2\sin^2\theta}\right)\,d\theta=\frac{1}{2}\int\limits_{0}^{2\pi}\log\left(1+2\left(\frac{R}{a\sin\theta}\right)^2\right)\,d\theta$$
The last integral looks more easy to bound hopefully.
A: We may assume $a>0$ without loss of generality.
We may decompose $D=[-R,R]^2$ as the union of $B^-,B^+$ and $C=D\setminus\left(B^+ \cup B^-\right)$, where 
$$B^-=\left\{(x,y)\in D : \left\|(x,y)-(-a,0)\right\|\leq\frac{a}{2}\right\} $$
$$B^+=\left\{(x,y)\in D : \left\|(x,y)-(a,0)\right\|\leq\frac{a}{2}\right\}. $$
If some point belongs to $B^+$ its distance from $(-a,0)$ is at least $\frac{3}{2}a$, hence
$$ \iint_{B^+}\frac{dx\,dy}{\sqrt{((x-a)^2+y^2)((x+a)^2+y^2)}}\leq \frac{2}{3a}\iint_{B^+}\frac{dx\,dy}{\sqrt{(x-a)^2+y^2}}=\frac{4\pi}{3}$$
where the last equality follows by switching to polar coordinates. By symmetry, the integral over $B^-$ is bounded by the same constant. Let $d^+=d^+(x,y)=\|(x,y)-(a,0)\|$, $d^-=d^-(x,y)=\|(x,y)-(-a,0)\|$ and
$ C^+ = C\cap\{x> 0\}$, $C^-=C\cap\{x<0\}$. We have:
$$ \iint_{C}\frac{dx\,dy}{d^+\cdot d^-}\leq\iint_{C}\frac{dx\,dy}{\min(d^+,d^-)^2}=2\iint_{C^+}\frac{dx\,dy}{d^+(x,y)^2}\leq 4\pi\int_{a/2}^{R\sqrt{2}}\frac{\rho}{\rho^2}\,d\rho= 4\pi\log\left(\frac{2R\sqrt{2}}{a}\right) $$
and since $\lim_{a\to 0^+}a\log(a)=0$, this proves that the wanted limit is zero.
