integral of a Kronecker product of exponentials

Question

Let $A\in\mathbb{R}^{n\times n}$. I do not know how to get a solution to the following integral:

$\int_{0}^{t}\left( e^{As}\otimes e^{As}\right) ds$

Answer 1

If $A$ is diagonalizable such that $A = U \cdot \Lambda \cdot U^{-1}$ for $\Lambda = {\rm diag}\{\lambda_1, \ldots, \lambda_n\}$, it's relatively easy. You then have ${\rm e}^{A s} = U \cdot {\rm e}^{\Lambda s} \cdot U^{-1}$, where ${\rm e}^{\Lambda s} = {\rm diag}\{{\rm e}^{\lambda_1 s}, \ldots, {\rm e}^{\lambda_n s}\}$.

Then, ${\rm e}^{A s} \otimes {\rm e}^{A s} = (U \otimes U) \cdot \bar{\Lambda} \cdot (U^{-1} \otimes U^{-1})$ where $\bar{\Lambda } = {\rm e}^{\Lambda s} \otimes {\rm e}^{\Lambda s} = {\rm diag}\{{\rm e}^{\lambda_is}{\rm e}^{\lambda_js}\}= {\rm diag}\{{\rm e}^{(\lambda_i+\lambda_j)s}\}$ for $i, j=1, 2, \ldots, n$.

The integral is then simply $$\int_0^t {\rm e}^{A s} \otimes {\rm e}^{A s}{\rm d}s = (U \otimes U) \cdot {\rm diag}\left\{\int_0^t {\rm e}^{(\lambda_i+\lambda_j)s}{\rm d}s\right\} (U^{-1} \otimes U^{-1}). $$

The inner integrals are now very simple, you get something like $\frac{1}{\lambda_i+\lambda_j}\left({\rm e}^{(\lambda_i+\lambda_j)t} - 1\right)$ for $\lambda_i+\lambda_j \neq 0$ and $t$ otherwise.