Current distribution of Hertz dipole NOT in origin of coordinate system

Your current distribution is incorrect (even for the origin), both unit-wise (if $I_0$ current) and meaning-wise. You are implying constant current distribution in the whole space.

For point-like z-aligned dipole at position $mathbf{r}_0$,

$$ mathbf{J}left(mathbf{r}right)=mathbf{hat{z}}I_0 l delta^{(3)}left(mathbf{r}-mathbf{r}_0right) $$

where $delta^{(3)}$ is the delta-function, and $l$ is some sort of a measure of how 'long' the dipole is. Normally you would pack $mathbf{hat{z}}I_0 l$=$mathbf{dot{p}}$ where $mathbf{p}$ is the dipole moment. Such 'packing' allows to hide the uncomfortable truth that if $lto0$, then $I_0toinfty$ for finite radiated power.

Alternatively you could use something like Heaviside function - $mathcal{H}$ and define:

$$ begin{align} Sleft(zetaright)=&mathcal{H}left(zeta+frac{1}{2}right) - mathcal{H}left(zeta-frac{1}{2}right) mathbf{J}left(mathbf{r}right)=&mathbf{hat{z}}I_0 Sleft(frac{z-z_0}{l}right) deltaleft(x-x_0right)deltaleft(y-y_0right)cosleft(frac{pi}{l}cdotleft(z-z_0right)right) end{align} $$

Where $l$ is now a 'proper' length of the dipole and $cos$ is needed for charge conservation.

I am implying harmonic time-dependence in both cases.

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Following comments. I think Balanis is not doing the best job in explaing here. Lets back up. In general you are trying to solve

$$ nabla^2 mathbf{A}+k^2mathbf{A}=-mumathbf{J} $$

I am assuming only electric and no magnetic current densities here. A solution in case of scattering boundary conditions (basically antenna radiates into free-space and no incoming waves) with no outside excitation is then:

$$ mathbf{A}left(mathbf{r}right)=frac{mu}{4pi}int d^3 r' mathbf{J}left(mathbf{r}'right)cdotfrac{expleft(-jkleft|mathbf{r}-mathbf{r}'right|right)}{left|mathbf{r}-mathbf{r}'right|} $$

This tells you the 'ammount' of vector potential $mathbf{A}$ you will see at position $mathbf{r}$ due to current density at position $mathbf{r}'$. You have to integrate over the whole space to pick up all the current density. In practice you often define current density to fit inside the volume $V$ in which case you write $int_V d^3 r'$. This is Balanis's equation Eq. 3-27, but with more explicit notation.

Now you see why spatial dependence of current density matters - it tells you how to limit your volume integral.

Next, how to represent a point-dipole antenna? Balanis talks about end-plates on the wires, Fig. 4-1. This allows him to define a consant current distribution in the wire with finite length. A proper expression for such distribution would be:

$$ begin{align} Sleft(zetaright)=&mathcal{H}left(zeta+frac{1}{2}right) - mathcal{H}left(zeta-frac{1}{2}right) mathbf{J}left(mathbf{r}right)=&mathbf{hat{z}}I_0 Sleft(frac{z-z_0}{l}right) deltaleft(x-x_0right)deltaleft(y-y_0right) end{align} $$

However Balanis does nothing interesting with this finite length of the antenna, so we can take the limit before computing the integral. This strays outside your question, so I will leave it for now.

Let us get back to solution and sub in the last current density. Using the properties of the delta-functions:

$$ begin{align} mathbf{A}left(mathbf{r}right)=&frac{mu}{4pi}int d^3 r' mathbf{J}left(mathbf{r}'right)cdotfrac{expleft(-jkleft|mathbf{r}-mathbf{r}'right|right)}{left|mathbf{r}-mathbf{r}'right|} =&mathbf{hat{z}}frac{mu I_0}{4pi}int^{z_0+l/2}_{z_0-l/2} dz'frac{expleft(-jksqrt{left(x-x_0right)^2+left(y-y_0right)^2+left(z-z'right)^2}right)}{sqrt{left(x-x_0right)^2+left(y-y_0right)^2+left(z-z'right)^2}} end{align} $$