Let $A\in\mathbb{R}^{n\times n}$ be a matrix whose real eigenvalues have negative real part, and $X=X^\top\in\mathbb{R}^{n\times n}$ be a positive semidefinite matrix, i.e., $X\succeq 0$. Consider the following matrix-valued function $$\tag{1}\label{1} F(X) = \left(\int_0^{\infty} e^{At} X e^{A^\top t} \mathrm{d} t\right)^{-1/2} X \left(\int_0^{\infty} e^{At} X e^{A^\top t} \mathrm{d} t\right)^{-1/2} $$ which maps positive (semi)definite matrices into a positive (semi)definite matrices. (Here $\cdot^{1/2}$ denotes the symmetric square root of a positive semidefinite matrix and $e^{\cdot}$ is the matrix exponential).
Note that \eqref{1} is a continuous function which is not defined for any $X\succeq 0$ such that $\int_0^{\infty} e^{At} X e^{A^\top t} \mathrm{d} t$ is singular. However, I wonder whether it is possible to define a continuous extension of $F(\cdot)$ in the set of positive semidefinite matrices.
In more formal terms, let $\bar{X}\succeq 0$ be such that $\int_0^{\infty} e^{At} \bar{X} e^{A^\top t} \mathrm{d} t$ is singular, and let $\{X_n\}_{n\ge 0}$, $X_n\succ 0$, be any sequence such that $\lim_{n\to \infty} X_n = \bar{X}$. Does $ \lim_{n\to \infty} F(X_n)$ exist and is finite?
My question is motivated by the special case of scalar matrices $A=\alpha I$, $\alpha<0$, for which it is easy to see that the above limit exists and is finite. For general $A$'s, however, it is not clear to me whether this limit still exists. Numerical simulations suggest that the answer is again in the affirmative, but I was not able to prove it.