Is the integral $x^* \left( \int_{0}^{\infty} e^{(B-A)^*t} (A^*A) e^{(B-A)t} dt \right) x$ unbounded when $\|A\| \to \infty$ in some matrix norm? Let $B \in \mathcal M(n \times n)$ be a fixed matrix and let 
$$S = \{A \in \mathcal M(n \times n; \mathbb C): \text{Re} \lambda_i(B-A) < 0 \text{ for } i = 1, \dots n\}.$$
Let us fix a matrix norm $\|\cdot\|_F$ (Here I use Frobenius norm for concreteness but it is irrelevant since the space is finite dimensional). It is clear the set $S$ is unbounded. I would like to know whether the following integral is unbounded
\begin{align*}
 \left(\sum_{j=1}^n x_j \right)^* \left( \int_{0}^{\infty} e^{(B-A)^*t} (A^*A) e^{(B-A)t} dt \right) \left(\sum_{j=1}^n x_j \right),
\end{align*}
as $S \ni \|A\|_F \to \infty$, where $\{x_j\}$ is some fixed spanning set of $\mathbb C^n$.
Intuitively, if $\|A\|_F \to \infty$, there must be unbounded entries and $A^* A$ must diverge to infinity in a faster manner. But I don't know how to bound the two matrix exponentials in terms of entries. The reason for me to believe the integral might be unbounded is: if we take one dimensional case, then it seems to be equivalent to
\begin{align*}
\int_{0}^{\infty} e^{-2nt} n^2 dt \to \infty \text{ as } n \to \infty.
\end{align*}
Is this true for the matrix integral? If so,
how should I make a precise argument?
 A: For simplicity, we write
$$P(A) = \int_{0}^{\infty} e^{(B-A)^*t}A^*Ae^{(B-A)t} \, dt.$$
Plugging $A = nI$ and applying the substitution $u = nt$, we obtain
$$ P(nI) = n\int_{0}^{\infty} e^{\frac{1}{n}B^*u}e^{\frac{1}{n}Bu}e^{-2u}\,du$$
Now notice that $ \left\| e^{\frac{1}{n}Bu} - I \right\| \leq e^{\frac{1}{n}\|B\|u} - 1 $ and the same inequality holds with $B^*$ in place of $B$, where, $\| \cdot \|$ denotes the operator norm. (Since any two matrix norms are equivalent, using $\|\cdot\|$ in place of $\|\cdot\|_F$ poses no hassle.) Then
\begin{align*}
\left\| \frac{1}{n}P(nI) - \frac{1}{2}I \right\|
&\leq \int_{0}^{\infty} \left\| e^{\frac{1}{n}B^*u}e^{\frac{1}{n}Bu} - I \right\| e^{-2u} \, du \\
&\leq \int_{0}^{\infty} \left( e^{\frac{1}{n}(\|B\|+\|B^*\|)u} - 1 \right) e^{-2u} \, du \\
&= \frac{1}{2-\frac{1}{n}(\|B\|+\|B^*\|)} - \frac{1}{2}
 = \mathcal{O}\left(\frac{1}{n}\right).
\end{align*}
Thus for any non-zero vector $x$,
\begin{align*}
\left| \langle x, P(nI) x \rangle - \frac{n}{2}\| x\|^2 \right|
\leq n \left\| \frac{1}{n}P(nI) - \frac{1}{2}I \right\| \|x\|^2
= \mathcal{O}(\|x\|^2)
\end{align*}
and therefore $ \langle x, P(nI) x \rangle \geq \left(\frac{n}{2}-\mathcal{O}(1)\right)\|x\|^2 $ is unbounded as $n\to\infty$.
