Let all eigenvalues of $A$ have a negative real part ( i.e. $A$ is stable ). Why does the following hold?

Question

Let all eigenvalues of $A$ have a negative real part ( i.e. $A$ is stable ). Why does the following hold? $$\int_{0}^{\infty} [Ae^{A\tau}BB^*e^{A^*\tau}+e^{A\tau}BB^*e^{A^*\tau}A^*] \, d\tau = \int_0^\infty d(e^{A\tau}BB^*e^{A^*\tau})$$ Please may you explain the meaning of $d(e^{A\tau}BB^*e^{A^*\tau})$ in the right hand side of the equation.

Many thanks,

Tri

Answer 1

Here I will give a partial answer without ruling out the possibility of finishing it later.

I am mildly skeptical of the equality because it seems to be neglecting a product rule that would say $\displaystyle \frac d {d\tau} \left( e^{A\tau}BB^*e^{A^*\tau}\right) = Ae^{A\tau}B B^* e^{A^*\tau} + e^{A\tau}BB^* A^* e^{A^*\tau}.$

But the other question is: What is the meaning of $d(e^{A\tau}BB^*e^{A^*\tau})$?

In the integral $\displaystyle \int_0^\infty d(e^{A\tau}BB^*e^{A^*\tau})$ one could wonder whether the variable that goes from $0$ to $\infty$ is $\tau$ or $A$ or $B$ or something else, but that is clear from the left side of the equality. The only meaning one can reasonably assign to this is that that integral is the total change in $e^{A\tau} B B^* e^{A^*\tau}$ as $\tau$ goes from $0$ to $\infty$, i.e. \begin{align} \int_0^\infty d\left(e^{A\tau} B B^* e^{A^*\tau}\right) & = \left( \lim_{\tau\to\infty} e^{A\tau} B B^* e^{A^*\tau} \right) - \left( e^{A0} B B^* e^{A^*0} \right) \\[6pt] & = \left( \lim_{\tau\to\infty} e^{A\tau} B B^* e^{A^*\tau} \right) - \left( B B^*\right). \end{align}