I wanted to calculate the magnetic field generated by a current which flows down the positive $z$axis,
hits the origin and then spreads out radially over the $xy$ plane, see figure 1.

Figure 1 
I started the calculation as follows. First I obtained the magnetic field generated by a semiinfinite wire. If the current comes out of the point $0$ (see figure 2), then the magnetic field in point $P$ is
[1]
\begin{equation}\label{eq:21070108a}
B_1(P) = \frac{\mu_0}{4 \pi\ s^2}\left( 1 + \frac{\vec I \cdot \vec P}{\vec I\ \vec P} \right) \vec I \times \vec P
\end{equation}
with $s$ the distance between $P$ and the wire.

Figure 2 
The magnetic field $B_2$ coming from all the currents in the $xy$ plane can be calculated by integrating \eqref{eq:21070108a}. The result is
\begin{equation}\label{eq:21070108b}
B_2(P) = \frac{\mu_0 I}{4 \pi}\left( \frac{z}{\vec P} \pm 1\right)\frac{1}{x^2 + y^2} \hat z \times \vec P
\end{equation}
The plus sign holds if $z < 0$, the minus sign if $z >0$. The vector $\hat z$ is the unit vector pointing in the direction of the $z$axis.
If I add $B_2$ to the field coming from the current flowing down the zaxis, some terms cancel and I get the final simple result.
\begin{align*}
B(P) & = \frac{\mu_0 I}{2 \pi s^2} \hat z \times \vec P\quad\text{if}\quad z>0\\
&=0 \quad\text{if}\quad z <0
\end{align*}
with $s$ the distance between $P$ and the $z$ axis. For $z>0$, $B$ is exactly equal to the magnetic field of an infinite wire! When I saw the final result, I realized it was easy to see that it was correct.
Indeed, for $z<0$ Maxwell's equations are trivially satisfied because the field is zero. For $z>0$ they are also satisfied because the field is equal to
magnetic field of an infinite wire. Finally, the currents in the $xy$ plane are also taken into account correctly. For example, take an Amperian loop as in figure 3.

Figure 3, $\gamma_1$ is above the $xy$ plane, $\gamma_3$ is below. 
Only the integral over $\gamma_1$ contributes. Also I know already that AmpĂ¨re's law is satisfied for $z > 0$ and the magnetic field is rotationally symmetric. Therefore
$$
\int_{\gamma} B \cdot dl = \mu_0 I \frac{\alpha}{2 \pi}
$$
with $\alpha$ the angle spanned by the curve $\gamma_1$
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