Example 10.2 of 3rd edition Griffith [electrodynamics] click here to read this question
So I thought to convert I into $\vec{J}$ as follows : $$\vec{J}(\vec{r},t)= I_o\theta(t)\delta(x)\delta(y)\hat{z}$$ Here $\theta(t)$ is Heaviside step function and $\delta(x)$ is just a Dirac delta function. And now we know that as $\rho$ is assumed as zero. So $\phi=0$, and now we know that the formula for vector potential is $$\vec{A}(\vec{r},t)=\dfrac{\mu_o}{4\pi}\int\int\int\frac{\vec{J}(\vec{r'},t_{retarded})}{\mathfrak{r}}d\tau'$$ If we take the above integral and solve then it doesn't reduce to this:- $$\vec{A}=\dfrac{\mu_o}{4\pi}\hat{z}\int{\frac{I}{\mathfrak{r}}dz}$$ as claimed by the author in the first step. click this to read the solution of the author . A small note for those might be familiar with electrodynamics. By 'I' I mean the electric current defined as $$I(t)\equiv\dfrac{d}{dt}q(t)$$ and here $$\vec{r}\equiv<x,y,z>$$ is position vector of any point in some given frame. Here $$\vec{r'}\equiv<x',y',z'>$$ be the position vector of the source charges and source currents. And let $$\vec{\mathfrak{r}}\equiv\vec{r}-\vec{r'}$$ be oistion vector defined with respect to the source point. And here $t_{retarted}\equiv t-\frac{\mathfrak{r}}{c}$
