I can't find a proof of the following fact: $$ \int_0^{+\infty} H\, \exp(tF)\, G \exp(-st)\ dt = H(s\,I - F)^{-1}G $$
where $H,F,G$ are constant matrices, $s$ is complex such that $s\,I - F$ is invertible.
First I assume that the sens of this integral should be understood component by component. I started to write complicated expression using the Gamma function: $$ \int_0^{+\infty} H\, \exp(tF)\, G e^{-st}\ dt =\sum_{k=0}^{\infty} H \frac{F^k}{k!} G\, \int_0^{\infty} t^k e^{-st}\ dt \\ =... $$ but with no conlusion
Because $H,G$ are constant (and sums and limits are linear), you have $$ \int_0^{+\infty} H\, \exp(tF)\, G \exp(-st)\ dt=H\,\left(\int_0^{+\infty} \, \exp(tF)\, \exp(-st)\ dt\right)\,G. $$ Now $$ e^{tF}e^{-st}=e^{-(sI-F)t} $$ As $sI-F$ is invertible, the antiderivative of $e^{-(sI-F)t}$ is $-(sI-F)^{-1}e^{-(sI-F)t}$. Then $$ \int_0^{m} \, \exp(tF)\, \exp(-st)\ dt=\left.\vphantom{\int}-(sI-F)^{-1}e^{-(sI-F)t}\right|_{t=0}^{t=m}=(sI-F)^{-1}(I-e^{-(sI-F)m}). $$ Now you want to take limit as $m\to\infty$. This doesn't always work. For instance, if $F=0$ and $s=-1$, the limit clearly doesn't exist. You need $F$ so satisfy some condition that guarantees that $\lim_{m\to\infty} e^{-(sI-F)m}=0$. This works for exampe if $sI-F$ is positive definite.