Consider a point charge $Q$ with a trajectory of $\textbf{s}(t)$ in frame $O$. The densities are: $$\rho(\textbf{x},t) = Q\delta^{(3)}(\textbf{x} - \textbf{s}(t))$$ $$\textbf{J}(\textbf{x},t) = Q\textbf{u}(t)\delta^{(3)}(\textbf{x} - \textbf{s}(t))$$ where $\textbf{u}(t) = \frac{d\textbf{s}(t)}{dt}$. By boosting from frame $O$ to $O'$ with arbitrary uniform velocity $\textbf{v} = c\beta$, show that the densities are covariant via explicit calculation.
From:
$$\rho'(\textbf{x}',t') = \gamma(\rho - \frac{1}{c}\beta\cdot\textbf{J}) = \gamma Q \delta^{(3)}(\textbf{x} - \textbf{s}(t)) (1 - \frac{1}{c}\beta \cdot \textbf{u}(t))$$
Thus, I can need to somehow show that it is the same as for covariance $$\rho'(\textbf{x}',t') = Q\delta^{(3)'}(\textbf{x}' - \textbf{s}'(t'))$$
However, I am not sure how to show the relation below: $$\gamma \delta^{(3)}(\textbf{x} - \textbf{s}(t)) (1 - \frac{1}{c}\beta \cdot \textbf{u}(t)) = \delta^{(3)'}(\textbf{x}' - \textbf{s}'(t'))$$ The usual transformation law of $\delta^{(3)'} = \gamma \delta^{(3)}$ doesn't seem to explain the $(1 - \frac{1}{c}\beta \cdot \textbf{u}(t))$ factor. I believe showing the relation for $\textbf{J}'$ would also follow similar argument.
