I have some questions regarding a special case of BCH-Formula, namely the case that
$$[X,Y]=sY$$ At first I have to show that the BCH-Formula reduces to $$\log(e^Xe^Y)=X+\frac{s}{1-e^{-s}}Y$$ via using the integral BCH-Formula and by showing that $e^Xe^{tY}$ and $e^{X+t\frac{s}{1-e^{-s}}Y}$ satisfy the same differential equation.
Furthermore that $e^Xe^Ye^{-X}=e^{e^sY}$ holds.
For the first part I tried several things with the integral formula. Although Wikipedia says it is evident from the integral formula, I do not see it. Is expanding $g(z)=\frac{z\log z}{z-1}$ to a specific order expedient?
For the differential equation approach I probably have to use the defintion of a derivative of a matrix exponential. $$\frac{d}{dt}e^{X(t)}=e^{X(t)}\left(\frac{I-e^{ad_{X(t)}}}{ad_{X(t)}}\frac{dX}{dt}\right)$$ But my attempt again came to nothing. Additionally I tried it with two matrices $$ X=\begin{pmatrix} s&0\\ 0&0 \end{pmatrix}$$ $$ Y=\begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}$$ that fulfill the bracket. I got $\log(e^Xe^Y)=\begin{pmatrix} s&s+1\\ 0&0 \end{pmatrix} $ And that is obviously not the same as $X+\frac{s}{1-e^{-s}}Y$.
In the last part I just have a small question. My attempt: $$e^Xe^Ye^{-X}=Ad_{e^X}Y=e^{ad_Xe^Y}=e^{ad_X(1+Y+Y^2/2+...)}=e^{sY+s^2/2Y+...}=e^{(e^{s}-1)Y}$$
How do I get rid of the $-1$ or where is my mistake?
This is a long list of questions.
Starting from the end, instead, $$e^Xe^Ye^{-X}=Ad_{e^X} ~e^Y=e^{ad_X} ~e^Y \\=e^{ad_X}(1+Y+Y^2/2+...)= 1+e^s Y+ e^{2s} Y^2/2+...= e^{e^s ~Y} ,$$ since ad$_X$ acting on Y is just multiplication scaling by s, so $ad_X ~ Y^n= ns Y^n$.
For your matrices $$ X=\begin{pmatrix} s&0\\ 0&0 \end{pmatrix}$$ $$ Y=\begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}$$ that fulfill the bracket, recall that logarithms of zero can lead to grief, so it is much easier to handle the exponentiated expressions, $$ e^X=\begin{pmatrix} e^s&0\\ 0&1 \end{pmatrix}, \qquad e^Y=\mathbb {I} +Y=\begin{pmatrix} 1&1\\ 0&1 \end{pmatrix}, \\ e^X e^Y=\begin{pmatrix} e^s&e^s\\ 0&1 \end{pmatrix}, \\ \exp \left ( X+\frac{s}{1-e^{-s}}Y \right ) =\exp \begin{pmatrix} s&se^s( e^s-1)\\ 0&0 \end{pmatrix} \\ = \mathbb {I} - \begin{pmatrix} 1&e^s/(e^s-1)\\ 0&0 \end{pmatrix} + e^s \begin{pmatrix} 1&e^s/(e^s-1)\\ 0&0 \end{pmatrix} \\ =\begin{pmatrix} e^s&e^s\\ 0&1 \end{pmatrix}, $$ alright.
For the formal parts you might consult my pedagogical crib notes, but I can expand this one.
For instance, the WP statement $$\log(e^Xe^Y)=X+\frac{s}{1-e^{-s}}Y$$ does, indeed, follow by inspection from the integral formula: ad$_Y$ has eigenvalue 0 when acting on Y, while ad$_X$ acting on Y keeps you in the space of Ys. So, subsequent actions of ad$_Y$ are also trivial; hence its exponential amounts to unity inside the argument of $\psi$, and the integration collapses to unity multiplying the remaining constant integrand. Now, acting on Y again, $\psi(e^{ad_X})~Y=\psi(e^{s})~Y={sY}/(1-e^{-s}) $.
The basic differential equation is likewise straightforward. On the one hand, $$ W\equiv e^X e^{tY} \qquad \Longrightarrow \qquad \partial_t W= WY. $$ On the other, $$ \tilde W \equiv e^V\equiv e^{X+ tY s/(1-e^{-s})}, $$ so that $$ \partial_t \tilde W= \tilde W ~ \frac {\mathbb{I}-e^{-ad_V}}{ad_V} \partial_t V =\tilde W ~ \frac {\mathbb{I}-e^{-ad_X}}{ad_X} ~\frac{s}{1-e^{-s}} ~Y = \tilde W Y. $$ This follows from $[V,Y]= [X,Y]=sY$ utilized above.