Integral involving matrix exponential

Question

Is there any way to simplify the integral $$ I = \int_{t_1}^{t_2}e^{\Lambda t} A e^{\Lambda t}\,dt $$ knowing that A is symmetric and Λ is a diagonal matrix?

Answer 1

Let $\lambda_1,\dots, \lambda_n$ denote the diagonal entries of $\Lambda$, and let $a_{ij}$ denote the entries of $A$. Note that we have $$ [e^{\Lambda t}A e^{\Lambda t}](i,j) = e^{(\lambda_i + \lambda_j)t} a_{ij} $$ Thus, integrating entrywise yields $$ I(i,j) = \frac{e^{(\lambda_i + \lambda_j)t_2} - e^{(\lambda_i + \lambda_j)t_1}}{\lambda_i + \lambda_j} $$ in the case that $\lambda_i \neq -\lambda_j$. If $\lambda_i = -\lambda_j$, then we end up with $I(i,j) = a_{ij}(t_2 - t_1)$.

Answer 2

Let $\Lambda\equiv\operatorname{diag}(\lambda_{1},\ldots,\lambda_{n})$ and $A\equiv(a_{ij})$. Then, $$ [e^{\Lambda t}Ae^{\Lambda t}]_{ij}=e^{(\lambda_{i}+\lambda_{j})t}a_{ij} $$ and hence $$ [I]_{ij}=a_{ij}\int_{t_{1}}^{t_{2}}e^{(\lambda_{i}+\lambda_{j})t}dt=a_{ij}\frac{e^{(\lambda_{i}+\lambda_{j})t_{2}}-e^{(\lambda_{i}+\lambda_{j})t_{1}}}{\lambda_{i}+\lambda_{j}}. $$

BTW, it seems like a dangerous idea to use the symbol $I$ to mean anything other than identity if there are matrices involved.