The retarded scalar potential produced by a charge distribution $\rho\left(t^{\prime}, \mathbf{x}^{\prime}\right)$ is

$$\varphi(t, \mathbf{x})=\frac{1}{4 \pi \epsilon_{0}} \int d^{3} x^{\prime} \frac{\rho\left(t-R, \mathbf{x}^{\prime}\right)}{R},$$

where $R=|\mathbf{R}|$ and $\mathbf{R}=\mathbf{x}-\mathbf{x}^{\prime}$. By use of an appropriate delta function rewrite the integral as an integral over both $d^{3} x^{\prime}$ and $d t^{\prime}$ involving $\rho\left(t^{\prime}, \mathbf{x}^{\prime}\right)$.

Now specialize to a point charge $q$ moving on a path $\mathbf{x}^{\prime}=\mathbf{x}_{0}\left(t^{\prime}\right)$ so that we may set

$$\rho\left(t^{\prime}, \mathbf{x}^{\prime}\right)=q \delta^{(3)}\left(\mathbf{x}^{\prime}-\mathbf{x}_{0}\left(t^{\prime}\right)\right) .$$

By performing the volume integral first obtain the Liénard-Wiechert potential

$$\varphi(t, \mathbf{x})=\frac{q}{4 \pi \epsilon_{0}} \frac{1}{\left(R^{*}-\mathbf{v} \cdot \mathbf{R}^{*}\right)},$$

where $\mathbf{R}^{*}$ and $\mathbf{v}$ should be specified.

Obtain the corresponding result for the magnetic potential.