How to evaluate $\int_0^{\infty}\int_0^{\infty}\left(\frac{e^{-\Lambda x}-e^{-\Lambda y}}{x-y}\right)^2 \,dx\,dy$ 
How to evaluate $$F(\Lambda)=\int_0^{\infty}\int_0^{\infty}\left(\frac{e^{-\Lambda x}-e^{-\Lambda y}}{x-y}\right)^2 \,dx\,dy$$ where $\Lambda$ is a positive real number?

I tried evaluating the innermost integral first,
$\displaystyle\int_{0}^{\infty}\left(\frac{e^{-\Lambda x}-e^{-\Lambda y}}{x-y}\right)^2 dx$ 
and found the solution is terms of the exponential integral $E_1(u)$ and I am not sure if that would help at all.

By symmetry, for any $\Lambda>0$ we have
$$\begin{eqnarray*} F(\Lambda)=\iint_{(0,+\infty)}\left(\frac{e^{-\Lambda x}-e^{-\Lambda y}}{x-y}\right)^2\,dx\,dy&=&2\int_{0}^{+\infty}\int_{0}^{x}\left(\frac{e^{-\Lambda x}-e^{-\Lambda y}}{x-y}\right)^2\,dy\,dx \\ &\stackrel{y\mapsto xz}{=}&2\int_{0}^{+\infty}\int_{0}^{1}\frac{\left(e^{-\Lambda x}-e^{-\Lambda xz}\right)^2}{x(1-z)^2}\,dz\,dx
\\ &\stackrel{x\mapsto w/\Lambda}{=}&2\int_{0}^{1}\int_{0}^{+\infty}\frac{\left(e^{-w}-e^{-wz}\right)^2}{w(1-z)^2}\,dw\,dz
\\ &\stackrel{\text{Frullani}}{=}&2\int_{0}^{1}\frac{2\log(1+z)-\log(4z)}{(1-z)^2}\,dz\end{eqnarray*}$$
hence $F(\Lambda)$ is constant and it equals
$$\begin{eqnarray*} F(\Lambda)&=&2\int_{0}^{1}\frac{2\log\left(1-\frac{u}{2}\right)-\log(1-u)}{u^2}\,du\\&=&2\int_{0}^{1}\sum_{n\geq 2}\left(\frac{1}{n}-\frac{2}{n 2^{n}}\right)u^{n-2}\,du\\&=& 2\sum_{n\geq 2}\frac{1}{n(n-1)}\left(1-\frac{2}{2^n}\right)=\color{blue}{2\log 2}. \end{eqnarray*}$$
