Special Case of Baker-Campbell-Hausdorff Formula 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?
 A: 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.
