I have an exercise that I can't seem to spit out. I want to use $$ \mathbf{I} - \int_0^\infty \mathbf{A} e^{-\mathbf{A}t}dt = \mathbf{0},\quad \forall\mathbf{A} \succ 0 $$ to prove that $$ \mathbf{B}^\dagger = \int_0^\infty e^{-\mathbf{B}^T\mathbf{B}t}\mathbf{B}^Tdt $$ where B is a full column rank matrix $\in \mathbb{R}^{m\times n}$.
This is what I got: Let $\mathbf{B}^T\mathbf{B} = A$ since $\mathbf{B}^T\mathbf{B}$ is positive definite by definition. Then using the first equation $$ \mathbf{I} - \int_0^\infty \mathbf{A} e^{-\mathbf{A}t}dt = \mathbf{0} $$ $$ \implies \mathbf{I} - \int_0^\infty \mathbf{B}^T\mathbf{B} e^{-\mathbf{B}^T\mathbf{B}t}dt = \mathbf{0} $$ $$ \implies \mathbf{I} = \int_0^\infty \mathbf{B}^T\mathbf{B} e^{-\mathbf{B}^T\mathbf{B}t}dt$$ $$ \implies \mathbf{B}^{\dagger}B = \int_0^\infty \mathbf{B}^T\mathbf{B} e^{-\mathbf{B}^T\mathbf{B}t}dt$$ since A is full column rank I didn't know where to go from here since I don't know if I can pull out a B from the integral and if so, which side it should go.
Any help appreciated