Given $A$ and $B$ are square matrices and $$\int_{0}^{+\infty}e^{-At}B B^T e^{-A^Tt}dt = I~~~~~~~~~(1)$$ Find the relationship between $A$ and $B$.
I want to get
$$\int_{0}^{+\infty}e^{-At}B B^T e^{-A^Tt}dt = (-A-A^T)^{-1}[e^{-At}B B^T e^{-A^Tt}]^{+\infty}_{0} ~~~~~~ (2)$$ If so, then I could get $$\int_{0}^{+\infty}e^{-At}B B^T e^{-A^Tt}dt= (-A-A^T)^{-1}(0-BB^T)= (A+A^T)^{-1}BB^T=I~~~~~~~(3)$$
But I don't know how to get rid of the integral, (i.e. I don't know how to get Equation (2) ).
Firstly, your integral (1) does not converge in general. An admissible hypothesis is: the eigenvalues of $A$ have a positive real part.
Secondly. Your calculation -for obtaining (2)- is false in general. Yet, it is valid when -for example- $AB=BA$ and $AA^T=A^TA$.
That is true is i) the integral (1) is a symmetric $\geq 0$ matrix.
EDIT. and ii) $\int_{0}^{+\infty}e^{-At}(A+A^T) e^{-A^Tt}dt = I$. Then the considered equality is equivalent to
$\int_{0}^{+\infty}e^{-At}(A+A^T-B B^T) e^{-A^Tt}dt = 0$.
A priori, that does not imply that $A+A^T=BB^T$. Yet, if we add the supplementary condition $A+A^T\leq BB^T$ or $A+A^T\geq BB^T$, then that works.