Why is the following a solution to the system? I have the following in my notes, but I can't remember how it works. Please help!
$\nabla^2\psi=0, \quad\psi\to 0\quad\text{as}\quad x^2+y^2\to\infty, \quad\psi (x,y,0)$ is continuous
Then by using Green's function, we get the solution to be 
$$\psi(x',y',z')={z'\over 2\pi}\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty [(x-x')^2+(y-y')^2+z'^2]^{-3\over 2}\psi(x,y,0)\,\,\,dxdy\;.$$
(This part I am sure about.)
The primed $x',y',z'$ are the variables introduced when using the Green's function $G(\vec{x};\vec{x'})$.
Why does this satisfy the boundary conditions? I am thinking that this solution is equivalent to
$$\psi(x,y,z)={z\over 2\pi}\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty [(x-x')^2+(y-y')^2+z^2]^{-3\over 2}\psi(x',y',0)\,\,\,dx'dy'\;.$$
But doesn't this imply that $\psi(x,y,0)\equiv 0? $ -- Not supposed to be true.
As an aside, are harmonic functions always spherically symmetrical?
Also, is it possible to actually evaluate that integral?
 A: I believe you're missing an absolute value on the factor $z$ outside the integral, or perhaps you've omitted a restriction to $z\ge0$. It can't be the solution the way you've written it, since this would be an odd function in $z$ and there's nothing in the problem that would break the symmetry between it and its negative.
You're right that these two ways of writing the solution are equivalent; you've merely renamed the variables.
The case $z=0$ has to be treated as a limit $z\to0$, since $z$ goes to $0$ while the integral goes to $\infty$. In the limit $z\to0$, the Green's function tends to a delta distribution (as it must to reproduce the boundary values at $z=0$). You can see this by integrating over it with constant $\psi(x,y,0)$; then the integral is always $1$, but the integrand is increasingly concentrated near $x=x'$, $y=y'$ as $z\to0$.
No, harmonic functions aren't always spherically symmetrical. For instance, the solution you've written down is a harmonic function, and it isn't spherically symmetrical.
