Context: This integral is part of the calculation of the step response to a linear system using the convolution equation, found in this book (pg. 150).
The part that is bugging me is the following integral:
$$ \int_{0}^{t} e^{A(t-\tau)} d \tau, $$ where $A$ is a matrix and $e^{A(t-\tau)}$ is the matrix exponential.
My approach was:
$$ \int_{0}^{t} e^{A(t-\tau)} d\tau = \int_{0}^{t} e^{At}e^{-A\tau} d\tau = e^{At}\int_{0}^{t} e^{-A\tau} d\tau = -e^{At}A^{-1}(e^{-At} - I), $$
where I used the property that $e^{A+B} = e^{A}e^{B}$ when $AB=BA$.
However, the author does the following: $$ \int_{0}^{t} e^{A(t-\tau)} d \tau = \int_{0}^{t} e^{A\sigma} d\sigma = A^{-1}(e^{At} - I). $$
I do not understand why my approach is wrong and why the author is allowed to do what he did.
We assume of course that $A$ is invertible. The author uses the substitution $\tau = t - \sigma$ , the identity $A^{-1} A = I$, the linearity of the integral and the property $\frac{\mathrm{d}}{\mathrm{d} t} \mathrm{e}^{t A} = A \mathrm{e}^{t A}$ of the matrix exponential to get \begin{align} \int \limits_0^t \mathrm{e}^{(t-\tau) A} \, \mathrm{d} \tau &= \int \limits_0^t \mathrm{e}^{\sigma A} \, \mathrm{d} \sigma = \int \limits_0^t I \, \mathrm{e}^{\sigma A} \, \mathrm{d} \sigma = \int \limits_0^t A^{-1} A \,\mathrm{e}^{\sigma A} \, \mathrm{d} \sigma \\ &= A^{-1} \int \limits_0^t A \, \mathrm{e}^{\sigma A} \, \mathrm{d} \sigma = A^{-1} \left[\mathrm{e}^{\sigma A} \right]_{\sigma = 0}^{\sigma=t} = A^{-1} \left(\mathrm{e}^{t A} - I\right) \, . \end{align} Note that since we have $ A^{-1} \mathrm{e}^{t A} = \mathrm{e}^{t A} A^{-1}$, your answer is not wrong, but completely equivalent: $$ - \mathrm{e}^{t A} A^{-1} \left(\mathrm{e}^{-t A} - I\right) = - A^{-1} \mathrm{e}^{t A} \left(\mathrm{e}^{-t A} - I\right)= A^{-1} \left(\mathrm{e}^{t A} - I\right) \, . $$