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enter image description hereLet 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

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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}

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    Thank you very much for your reply Michael. You were right I made a big mistake when copying the equation. My apologies. I have corrected this belatedly. – 2017-01-25
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    Regarding the first point, you might write $e^{A^*\tau}=[e^{A\tau}]^*$. So, $de^{A^*\tau}=d[e^{A\tau}]^*=[Ae^{A\tau}]^*d\tau=e^{A^*\tau}A^*d\tau$. This seems to agree with the expression in the question. – 2017-01-27