I tried applying the following relation
$$ \vec{A} = \frac{\mu_0}{4 \pi} \int_\Omega \frac{\vec{J}}{\vert \vec{r} - \vec{r}^\prime \vert} dV$$
To find the magnetic vector potential on all space of a current loop of radius R carrying a current I, lying on the xy plane. I will be using spherical coordinates, where $(r^\prime, \theta, \phi)$ are the coordinates used for the configuration and $(r, \psi, 0)$ where the field is being evaluated. We have
$$ \vec{J} = I \frac{\delta (r^\prime - R) \delta( \theta - \frac{\pi}{2}) }{r^\prime} \hat{\phi}$$ For the current density and
$$ {\vert \vec{r} - \vec{r}^\prime \vert}^2 = r \prime^2 + r^2 -2r r^\prime (\sin\theta \cos\phi \sin \psi + \cos \psi \cos \theta)$$
Where $\psi$ is our angle with the z axis. Now (writing $\hat{\phi}$ in terms of the cartesian basis) our integral is
$$ \vec{A}(r, \psi) = \frac{\mu_0}{4 \pi} I \int_\Omega \frac{\delta (r^\prime - R) \delta( \theta - \frac{\pi}{2}) (\cos \phi \hat{j} - \sin \phi \hat{i})}{\sqrt{ r^{\prime 2} + r^2 -2r r^\prime (\sin\theta \cos\phi \sin \psi + \cos \psi \cos \theta)}} r^\prime \sin \theta dr d\theta d\phi$$
Evaluating $r$ and $\theta$ over $[0, \infty)$ and $[0, \pi]$ respectively
$$ = \frac{\mu_0}{4 \pi} I R \int_0^{2\pi} \frac{\cos \phi \hat{j} - \sin \phi \hat{i}}{\sqrt{r^2 + R^2 -2r R (\cos\phi \sin \psi)}} d\phi $$
Since this is pen and paper to solve the integral I replaced the denominator with its multipole expansion to the first power
$$ = \frac{\mu_0}{4 \pi} I R \int_0^{2\pi} \frac{1}{r} (1 - \frac{1}{2} (\frac{R}{r})^2 + \frac{R}{r} \cos \phi \sin \psi) (\cos \phi \hat{j} - \sin \phi \hat{i}) d\phi$$
Whiteout risking boring the reader, this integral evaluates to $0$ due to the $\phi$ terms. I would expect this to happen when $\phi=0$ due to the azimuthal symmetries on the loop's axis, but I don't understand why is it $0$ outside the axis, and can't find my computational error. I know this can't be explained by the choice of gauge, since we obviously know a current loop does produce a magnetic field and the gauge can't cancel the entire potential.
