I have a relatively specific problem. Consider a linear system, $\dot{x}(t) = Ax(t)+Bu(t)$, where $A \in \mathbb{R}^{n \times n}$ and $B \in \mathbb{R}^{n \times 1}$ are the constant matrix and constant vector that defines the system.
The closed-form for a linear system with a constant control input $u(t) := u_k, \forall t\in [0, T]$ has the following form: $x(t) = e^{At}x(0) + \int_{0}^{T} e^{A(t-\tau)} d\tau Bu$. If $A$ is non-singular (invertible), then $\int_{0}^{T} e^{A(t-\tau)} d\tau $ can be rewrite to $A^{-1}(e^{AT}-I)$, where $I$ is an identity matrix with appropriate dimension. At this point, I can rewrite $x(t) = e^{At}x(0) + A^{-1}(e^{AT}-I) Bu$, which does not have the integral.
But a lot of systems in real-life have a singular matrix $A$ for its dynamics ($A$ is not invertible ), such as a double integrators with $A = [0,1;0,0]$. In this case, are there any alternatives for me to write its closed-form solution that does not contain integral?
You can use a trick that is also often used to obtain a zero-order-hold discetization. Namely
$$ e^{\begin{bmatrix}A & B \\ 0 & 0\end{bmatrix}t} = \begin{bmatrix}e^{A\,t} & \int_0^t e^{A(t-\tau)} d\tau\,B \\ 0 & I\end{bmatrix}. $$
If you want to verify this for yourself you can try to expand the matrix exponential, which can be defined as
$$ e^M = I + M + \tfrac{1}{2}M^2 + \tfrac{1}{3!}M^3 + \cdots. $$