= Solution
The scalar contraction of the <electromagnetic field tensor> is
$$
F_{\mu\nu}F^{\mu\nu}=2(B^2-E^2).
$$
With orientation $\epsilon^{0123}=+1$, define the <dual electromagnetic field tensor> by $\widetilde F^{\mu\nu}=\tfrac12\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}$. Direct substitution gives
$$
\boxed{E^2-B^2=-\frac12F_{\mu\nu}F^{\mu\nu},\qquad
\mathbf E\cdot\mathbf B=-\frac14F_{\mu\nu}\widetilde F^{\mu\nu}.}
$$
The first is a <Lorentz scalar>; the second is a <pseudoscalar>, invariant under proper <Lorentz transformations> such as boosts and rotations. Under spatial inversion $\mathbf E$ changes sign while $\mathbf B$ does not, so $\mathbf E\cdot\mathbf B$ reverses sign. For example, $\mathbf E=\mathbf B=\widehat{\mathbf z}$ has product $+1$ before inversion and $-1$ afterwards. Thus the unrestricted statement including improper transformations needs this qualification.
For the spacetime integral, assume a globally regular <electromagnetic four-potential> on ordinary Minkowski space and sufficient boundary decay. Using $\nabla\cdot\mathbf B=0$ and $\partial_t\mathbf B=-\nabla\times\mathbf E$,
$$
\begin{aligned}
\partial_t(\mathbf A\cdot\mathbf B)
&=(-\mathbf E-\nabla\phi)\cdot\mathbf B-\mathbf A\cdot(\nabla\times\mathbf E)\\
&=-2\mathbf E\cdot\mathbf B-\nabla\cdot(\phi\mathbf B+\mathbf E\times\mathbf A).
\end{aligned}
$$
Consequently
$$
\boxed{\mathbf E\cdot\mathbf B=-\frac12\partial_t(\mathbf A\cdot\mathbf B)
-\frac12\nabla\cdot(\phi\mathbf B+\mathbf E\times\mathbf A).}
$$
The <magnetic helicity boundary term> controls the integral. If its surface flux vanishes and the <magnetic helicity> $H_B(t)=\int\mathbf A\cdot\mathbf B\,d^3x$ vanishes at both temporal ends, then
$$
\boxed{\int_{\mathbb R^4}\mathbf E\cdot\mathbf B\,d^4x
=-\frac12[H_B(+\infty)-H_B(-\infty)]=0.}
$$
Compactly supported smooth potentials are one sufficient version of the intended falloff assumption. The local identity is exact, but \b[vanishing fields at spatial infinity alone is insufficient]. For a concrete counterexample, take $\phi=0$ and, in dimensionless coordinates,
$$
\mathbf A=a(t)e^{-(x^2+y^2+z^2)}(-y,x,\lambda),\qquad \lambda\ne0,
$$
where $a$ changes smoothly from zero to one and $\dot a$ has compact temporal support. Both <electric fields> and <magnetic fields> vanish rapidly at spatial infinity, but
$$
H_B(t)=2\lambda\left(\frac\pi2\right)^{3/2}a(t)^2,
\qquad
\int\mathbf E\cdot\mathbf B\,d^4x
=-\lambda\left(\frac\pi2\right)^{3/2}\ne0.
$$
The counterexample leaves a final magnetic field and therefore fails the extra temporal boundary condition. It shows exactly why the printed integral cannot follow from a merely spatial reading of “at infinity”; vanishing of the full boundary term is the needed hypothesis.
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