If u(t) is the unit step function, then the region of convergence (ROC) of the Laplace transform of the signal.

$x(t)=e^{t^2}[u(t-1)-u(t-10)]$

The correct option is:
$-\mathrm{\infty }<\mathrm{Re}(s)<\mathrm{\infty }$

Step-by-Step Explanation:

1. Understand the Signal:
We are given the signal:
$x(t)=e^{t^2}[u(t-1)-u(t-10)]$

2. Analyze the Time Interval:
The term $[u(t-1)-u(t-10)]$ is a rectangular pulse that is equal to 1 for $1\le t<10$, and 0 elsewhere.
Therefore, the signal $x(t)$ is non-zero only within a finite time interval:
$x(t)=\{\begin{array}{ll}{e}^{t^2}& \text{for}1\le t<10\\ 0& \text{otherwise}\end{array}$

3. Determine the ROC of a Finite-Duration Signal:
A signal that is non-zero only over a finite time interval $[t_1,t_2]$ is called a finite-duration signal.
If a finite-duration signal is absolutely integrable (meaning it does not grow to infinity within its bounded domain), its Laplace transform converges for all values of $s$ in the complex plane.
Here, the function $e^{t^2}$ is continuous and bounded on the closed interval $[1,10]$. Thus, the integral for the Laplace transform:
$X(s)={\int }_1^{10}$
is guaranteed to converge to a finite value for any complex number $s$, because the limits of integration are finite and the integrand is bounded.
Consequently, the Region of Convergence (ROC) is the entire s-plane, which is represented by:
$-\mathrm{\infty }<\mathrm{Re}(s)<\mathrm{\infty }$