It's well known that the derivative of the matrix exponential in direction $V$ is given by
\begin{align}
D_V(t,A)&=\int_0^t \exp((t-s)A)V\exp(sA)ds
\end{align}
In [1] they state that one can also write it as (eqn. 104)
\begin{align}
D_V(t,A)&=\int_0^t \exp(sA)V\exp((t-s)A)ds
\end{align}
Why are these equivalent?
[1] Najfeld, Igor, and Timothy F. Havel. "Derivatives of the matrix exponential and their computation." Advances in applied mathematics 16, no. 3 (1995): 321-375.
Best Answer
Begin with the first integral formula, and apply the substitution $\sigma = t - s$. We find that $$ \int_0^t \exp((t-s)A)V\exp(sA)ds = \\ \int_{\sigma(0)}^{\sigma(t)} \exp(\sigma A) V \exp((t - \sigma )A)\cdot (-1)d \sigma =\\ -\int_t^0 \exp(\sigma A)V \exp((t - \sigma A))\,d\sigma =\\ \int_0^t \exp(\sigma A)V \exp((t - \sigma A))\,d\sigma, $$ which is the desired formula.