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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 ${(M(x)-1)y}/{(M(x)(1-y)+y)}*L$, $M(x)=Ad(e^Ke^{xL})$, on $[0,1]²$, the integral over $y$ treated as a Cauchy principal value.

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 ${(M(x)-1)y}/{(M(x)(1-y)+y)}*L$, $M(x)=\operatorname{Ad}(e^Ke^{xL})$, on $[0,1]^2$, the integral over $y$ treated as a Cauchy principal value.

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 ${(M(x)-1)y}/{(M(x)(1-y)+y)}*L$, $M(x)=Ad(e^Ke^{xL})$, on $[0,1]²$.

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 ${(M(x)-1)y}/{(M(x)(1-y)+y)}*L$, $M(x)=Ad(e^Ke^{xL})$, on $[0,1]²$, the integral over $y$ treated as a Cauchy principal value.

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 ${(M(x)-1)y}/{(M(x)(1-y)+y)}*L$, $M(x)=Ad(e^Ke^{xL})$, with respect first to $y$ (as a Cauchy principal value w.l.o.g.) and then with respect to $x$, on $[0,1]^{2}$.

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 ${(M(x)-1)y}/{(M(x)(1-y)+y)}*L$, $M(x)=Ad(e^Ke^{xL})$, on $[0,1]²$.