Consider $A$ a symetric matrix such that $\mu(A) = \max\{Re(\lambda). \lambda \in \sigma(A)\} < 0$.
Justify that $B = \int_0^{\infty} e^{A^*t} e^{At} dt$ is well-defined.
I read that $\mu(A) < 0$ justifies well-definiteness of $B$. These are my remarks:
$\|\int_0^{t} e^{A^*t} e^{At} dt\| \le \int_{0}^t \| e^{A^*t} e^{At}\| dt \le C^2 (1+|t|)^{2d+1} e^{2 \mu(A) t} \to 0$
Here I used @SangchulLee bound. However, with this bound I'm getting that the integral converges to $0$ and this doesn't seem to make sense since $B$ should be positive definite (see related question).
Related
Show that the for lyapunov equation $A^TQ+QA=-I$, the matrix $Q$ is positive definite.
Your mistake is similar to saying that the gamma function is $0$, maybe you applied a sup bound incorrectly? You can just compute the upper bound. Write $r=2|\mu(A)|>0$. Then continuing from Sangchul's bound, \begin{align} \left\|\int_0^t e^{A^*s}e^{As}ds\right\| &\le C^2\int_0^t(1+s^d)^2e^{2\mu(A)s}ds \\&= C^2\int_0^t(1+s^d)^2e^{-rs}ds \\ &=C^2r^{-1}\int_0^{rt} (1+(x/r)^d)^2e^{-x}dx\\ &=C^2r^{-1}\int_0^{rt}e^{-x}dx +2C^2r^{-d-1}\int_0^{rt} x^de^{-x}dx \\&\quad + C^2r^{-2d-1}\int_0^{rt} x^{2d}e^{-x}dx \\ &\xrightarrow[t\to\infty]{} C^2r^{-1} + 2C^2r^{-d-1}\Gamma(d+1) \\&\qquad +C^2r^{-2d-1}\Gamma(2d+1) \in (0,\infty) \end{align}