Gauge Invariant terms of Lagrangian for Electromagnetism Besides the usual EM Lagrangian $\mathcal{L} = -\frac{1}{4}F^{\mu \nu}F_{\mu \nu}$, we can add an additional term $\mathcal{L'} = \epsilon_{\mu \nu \rho \sigma }F^{\mu \nu}F^{\rho \sigma} = -8 \vec{E} \cdot \vec{B} $. Then, I want to show that adding $\mathcal{L'}$ does not affect the Maxwell's equations. In order to prove it, I wanna show that $ \int d^{4}x \mathcal{L'} = C$, where $C$ is a constant. If  $ \int d^{4}x \mathcal{L'} = C$ by least action principle, the Maxwell equations are unchanged. By Attempt is as follow:
$\textbf{Attempt}$:
\begin{equation}
\int d^{4}x \mathcal{L'} = -8  \int d^{4}x \vec{E} \cdot \vec{B}
\end{equation}
Recall that $\vec{B} = \nabla \times \vec{A}$, where $\vec{A}$ is the vector potential. In order to evaluate the integration, we use the following identity:
\begin{equation}
\nabla \cdot ( F \times G) = G\cdot (\nabla \times F) - F \cdot(\nabla \times G) 
\end{equation}
In here, we replace  $F \rightarrow E$ and $G \rightarrow A$, we can obtain the following:
\begin{equation}
\nabla \cdot( \vec{E} \times \vec{A}) = \vec{A} \cdot ( \nabla \times \vec{E}) - E \cdot (\nabla \times \vec{A}) 
\end{equation}
By applying the integration on both sides:
\begin{equation}
\int_{\Omega} d^{4}x \nabla \cdot( \vec{E} \times \vec{A}) = \int_{\Omega} d^{4}x \Big( \vec{A} \cdot ( \nabla \times \vec{E}) \Big)  -  \int_{\Omega} d^{4}x  \Big( E \cdot (\nabla \times \vec{A})  \Big)
\end{equation}
By Divergence theorem, we can reduce the left-handed side:
\begin{equation}
\int_{\Omega} d^{4}x \nabla \cdot( \vec{E} \times \vec{A}) = \int_{\partial \Omega} dS ~\hat{n} \cdot ( \vec{E} \times \vec{A}) 
\end{equation}
in here, I assume that the boundary  $\partial \Omega$ is very far away from the origin and both $\vec{E} =\vec{A} = 0 $ at the boundary. Besides, $\nabla \times \vec{E} = 0$ therefore $\int_{\Omega} d^{4}x  \Big( E \cdot (\nabla \times \vec{A})  \Big) = 0  \rightarrow \int d^{4}x \mathcal{L'} = 0 $. Hence, the Maxwell equations are unchanged. This is my attempted proof. However, I am not sure whether I evaluate the integration  $\int_{\partial \Omega} dS ~\hat{n} \cdot ( \vec{E} \times \vec{A}) $ and $\nabla \times \vec{E} =0$  are correct. Could anyone point out that whether my proof is correct? I appreciate any comment.
 A: Your approach works, but as @bolbteppa notes it's easier to work covariantly. Either way, it's a "because boundary terms vanish" argument.
Let's repeatedly use antisymmetries. Since $\partial_\mu(\epsilon^{\mu\nu\rho\sigma}\partial_\rho A_\sigma)=\epsilon^{\mu\nu\rho\sigma}\partial_\mu\partial_\rho A_\sigma=0$,$$\partial_\mu(\epsilon^{\mu\nu\rho\sigma}A_\nu\partial_\rho A_\sigma)=\epsilon^{\mu\nu\rho\sigma}\partial_\mu A_\nu\partial_\rho A_\sigma=\tfrac14\epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma},$$so $\int\mathcal{L}^\prime d^4x\propto\int\partial_\mu(\epsilon^{\mu\nu\rho\sigma}A_\nu\partial_\rho A_\sigma) d^4x$ is a boundary term. It's also worth noting how direct work with Euler-Lagrange equations looks:$$\frac{\partial\mathcal{L^\prime}}{\partial\partial_\mu A_\nu}\propto\epsilon^{\mu\nu\rho\sigma}\partial_\rho A_\sigma\implies\frac{\partial\mathcal{L^\prime}}{\partial A_\nu}\stackrel{\text{on-shell}}{\propto}\partial_\mu(\epsilon^{\mu\nu\rho\sigma}\partial_\rho A_\sigma)=0,$$i.e. the ELE is still $\partial_\mu F^{\mu\nu}=0$.
