Thank you Carlo! I didn't know about the existence of the last work, which seems quite general for the $2\times2$ case, and solves the problem at least in this case.

Thank you for your answer. The first expression you get is the one that it is obtained by changing the sum by an integral in the solution given by M. Alekseyew. However, I didn't know about the existence of these polylogarithms, which are useful for simplifying the expressions. If you have some figures I would like to see them (if possible)

Do you think it would be a good approach to the nice exact result you got?

Thank you for the answer, I found it so useful. At the end I think that the series you get before defining the Eularian numbers can be sumed up by converting the sum into an integral:\begin{equation}\sum_{k=1}^\infty k^{j-1}/\alpha^k\to \int_{1}^{\infty}dx\;x^{j-1} e^{-\alpha x}\end{equation}