Does $ \int_{X}^{\infty} \int_{X}^{\infty} \min\left\{1, \frac{1}{|x_1 - x_2|}\right\} \frac{1}{(\max \{x_1, x_2\})^2} d x_1 d x_2 $ converge? Let $X>0$ be a fixed real positive number. 
I am trying to figure out if the following integral converges or diverges. 
$$
\int_{X}^{\infty} \int_{X}^{\infty} \min\left\{1, \frac{1}{|x_1 - x_2|}\right\}  \frac{1}{(\max \{x_1, x_2\})^2} d x_1 d x_2.
$$
Any input would be appreciated. Thank you! 
 A: The inner integral gives
$$
\int_X^\infty\dots\,\mathrm{d}x_1=
\begin{cases}
\frac{x_2 + \log(x_2 + 1) - X}{x_2^2} & X<x_2\leq 1+X\\
\frac{\log(x_2 + 1) + \log(x_2 - X) + 1}{x_2^2} & x_2>1+X
\end{cases}
$$
so then the outer integral converges.
A: i) Along the line $x=y$, the integrand is close to $\sum_{n=1}^\infty\ \frac{1}{n^2}$
ii) Consider a line $y=x-n$ so that $$
\frac{1}{|y-x|}=\frac{1}{n},\ \frac{1}{{\rm max}\ \{x,y=x-n=i\}^2}
=\frac{1}{(n+i)^2}$$
Hence $2\sum_{n=1}^\infty \frac{1}{n}\sum_{i=0}^\infty\
\frac{1}{(n+i)^2} =\ast $
Here $\int_n^\infty\ \frac{1}{x^2}\ dx =\frac{1}{n}$ so that
$\sum_{i=0}^\infty\ \frac{1}{(n+i)^2}$ is close to $\frac{1}{n}$. So
$\ast$ is convergent so that so is the given integral.
A: Since the integrand is symmetric WRT $x_1$ and $x_2,$ we can restrict to $X\leq x_2\leq x_1$ and then double it, so the integral is equal to:
$$\begin{align}I(X)&=2\int_{X}^{\infty}\int_{X}^{x_1}\min\left(1,\frac{1}{x_1-x_2}\right)\frac{1}{x_1^2}\,dx_2\,dx_1\\
&=2\int_{X}^{\infty}\frac{1}{x_1^2}\int_{0}^{x_1-X}\min\left(1,\frac{1}{u_2}\right)\,du_2 \,dx_1
\end{align}$$
where $u_2=x_1-x_2.$
Now, $$J(X,x_1)=\int_{0}^{x_1-X}\min\left(1,\frac{1}{u_2}\right)\,du_2=\begin{cases}x_1-X& X\leq x_1\leq X+1\\
1+\log(x_1-X)&x_1>X+1\end{cases}$$
So:
$$\begin{align}I(X)&=2\int_{X}^{\infty} \frac{J(X,x_1)}{x_1^2}\,dx_1\\
&=2\left(\int_{X}^{X+1}\frac{(x_1-X)}{x_1^2}\,dx_1 +\int_{X+1}^{\infty}\frac{1+\log(x_1-X)}{x_1^2}\,dx_1\right)\\
&=2\left(\log(X+1)-\log(X) +\frac{X}{X+1}-1+\frac{1}{X+1}+\int_{X+1}^{\infty}\frac{\log(x_1-X)}{x_1^2}\,dx_1\right)\\
&=2\left(\log(X+1)-\log(X) +\int_{X+1}^{\infty}\frac{\log(x_1-X)}{x_1^2}\,dx_1\right)
\end{align}$$
Letting $u=\log(x_1-X)$ and $dv=\frac{dx_1}{x_1^2}$ then $v=\frac{-1}{x_1}$ and $du=\frac{dx_1}{x_1-X}.$ Then:
$$\begin{align}\int_{X+1}^{\infty}\frac{\log(x_1-X)}{x_1^2}\,dx_1 &= \int_{X+1}^{\infty}\frac{dx_1}{x_1(x_1-X)}\\
&=\frac{1}{X}\int_{X+1}^{\infty}\left(\frac{1}{x_1-X}-\frac{1}{x_1}\right)\,dx_1\\
&=\frac{1}X \left(\log(X+1)\right)
\end{align}$$
So the final result is:
$$I(X)=2\left(\log(X+1)-\log(X)+\frac{\log(X+1)}{X}\right)$$
