We fix $\epsilon>0$ and consider the open ball in $\mathbb{R}^3$ $$B_\epsilon=\left\{x\in \mathbb{R}^3:\lVert \mathbf{x} \rVert<{\epsilon}\right\}.$$
If $\mathbf{F}$ is a vector field of class $C^1$ on the closure of $B_\epsilon$, then by the Divergence Theorem we know that $$\int_{B_\epsilon} \nabla\cdot \mathbf{F}=\int_{\partial B_\epsilon}\mathbf{F}\cdot \mathbf{n}\qquad\qquad\qquad[1] $$where $\mathbf{n}$ is the outward pointing unit normal field of the boundary $\partial B_\epsilon=\left\{x\in \mathbb{R}^3:\lVert \mathbf{x} \rVert={\epsilon}\right\}$. The left side of the previous is a volume integral, the right side is a surface integral.
I would like to know if $[1]$ holds on the unbounded domain $D_\epsilon:={\overline{B_\epsilon}}^c$. More precisely:
Lets define $$D_\epsilon=\left\{x\in \mathbb{R}^3:\lVert \mathbf{x} \rVert>{\epsilon}\right\},$$ and suppose $\mathbf{F}$ is a vector field of class $C^1$ on $\overline{D_\epsilon}=\left\{x\in \mathbb{R}^3:\lVert \mathbf{x} \rVert\geq{\epsilon}\right\}$. Here, obviously, $\partial D_\epsilon =\partial B_\epsilon$.
If the improper Riemann integral of $\nabla \cdot \mathbf{F}$ on $D_\epsilon$ is convergent, can we state that $$\int_{D_\epsilon} \nabla\cdot \mathbf{F}=\int_{\partial D_\epsilon}\mathbf{F}\cdot \mathbf{n}\qquad ?\qquad\qquad\qquad[2] $$
Any hint for a proving/disproving $[2]$ would be really appreciated.