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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.

mlstudent
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1 Answers1

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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.

Ben Grossmann
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  • Thanks. I was working through it myself and got to the third line and got confused. What does an integral from 0 to t mean and why can we invert it to get its negative? – mlstudent Oct 12 '21 at 03:34
  • @mlstudent I'm not sure how to answer your question about what the integral "means". In general, $\int_a^b f(x),dx = -\int_b^a f(x),dx$; switching the limits of an integral multiplies it by $-1$. – Ben Grossmann Oct 12 '21 at 03:36
  • Actually I found a discussion here https://math.stackexchange.com/questions/1577675/why-change-the-sign-of-the-integral-when-switching-the-limits-of-integration. – mlstudent Oct 12 '21 at 03:39
  • @mlstudent This statement holds for real-valued integrals, perhaps it would be helpful to convince yourself of that simpler statement – Ben Grossmann Oct 12 '21 at 03:39
  • Sorry when I said from 0 to t I meant from t to 0. – mlstudent Oct 12 '21 at 03:42