Answering my own question.

Comes trivially from the BCH-D integral expression; "$ln(e^Ke^L)-K-L$" being equal to the iterated integral of ${(M(x)-1)y}/{(M(x)(1-y)+y)}*L$, $M(x)=e^Ke^{xL}$, with respect first to $y$ (as a Cauchy principal value w.l.o.g.) and then with respect to $x$, on $[0,1]^{2}$.