# integral of a Kronecker product of exponentials

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

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

• Should $B$ appear somewhere in your integral? – John Hughes Apr 5 at 11:26
• Is $A$ diagonalizable? – Florian Apr 5 at 14:13

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.