96 Exponential and Inverse, concluded

$$(sI-A){\int }_0^R\exp (t(A-sI))𝑑t={\left[-\exp (t(A-sI))\right]}_0^R=I-\exp R(A-sI);$$

by the assumption on $s$, we can take the limit as $R\to \mathrm{\infty }$ to obtain

$$(sI-A){\int }_0^{\mathrm{\infty }}\exp (t(A-sI))𝑑t=I.$$

Hence $sI-A$ is invertible. The previous formula is often called the resolvent, as it helps solve equations.