How to find an integral kernel for poisson's equation in the upper half plane In our lecture we have shown that $\forall f \in L^2(\mathbb{R}^n_+) $ there is a unique $ u $ in the Sobolev space $ H^2(\mathbb{R}^n_+) $ satisfying $ -\Delta u = f. $
Now in our exercise sheet we are asked to show that there is an integral kernel $ \Phi $
such that $ u(x) = \int_{\mathbb{R}^n} \ \Phi (x-y) \ f(y) \ dy $.
Wikipedia tells me that there is an integral kernel and that it is of the form 
\begin{equation*} \Phi(x) \ = \ const. \ \cdot \ \frac{x_n}{({\sum_{i = 1}^{n} x_i^2})^{n/2}} \end{equation*}
So now to my question: 
How can you show that this is indeed an integral kernel for poisson's equation? In particular, how can you differentiate under the integral sign and "take the Laplacian" of $ \Phi $ at $ x - y = 0 $ ? 
Moreover, do you know a priori that there has to be such an integral kernel?
Thanks a lot in advance, I would really appreciate your help!
Best regards 
Phil
 A: it's very late and probably you know the answer already, but in case others are wondering about the same question, I thought I'd write a few lines.
If you want to solve $-\Delta u=f$ on $\mathbb{R}^n$, then $u(x)=\int_{\mathbb{R}^n}\Phi(x-y)f(y)dy$, where 
$$\Phi(x)=
\left\{
 \begin{array}{ll}
  -\frac{1}{2\pi}\mathrm{log}(|x|)  & \mbox{if } n = 2 \\
  \frac{1}{n(n-2)\alpha(n)}\frac{1}{|x|^{n-2}} & \mbox{if } n \geq 3
 \end{array}
\right.$$
A possible reference is Chapter 2.2 of Evans: Partial Differential Equations (in particular pp. 20-25). There you find a detailed proof of this statement in the case that $f \in C_{c}^2(\mathbb{R}^n)$, that is, $f$ is twice continuosly differentiable with compact support. In particular, Evans takes care about your (justified) worries about "differentiating through the integral sign". On the other hand, I don't know if there is any argument for the existence of this integral kernel a priori.
However, as Andrew pointed out already, the integral kernel you have written down is the Poisson kernel for $\mathbb{R}_{+}^{n}$, which is designed to solve Laplace's equation ($\Delta u =0$) on $\mathbb{R}_{+}^{n}$, and NOT, as its name might suggest, Poisson's equation ($-\Delta u =f$), which you want to solve (supposedly). If you want to solve Laplace's equation on $\mathbb{R}_{+}^{n}$ with boundary condition $u=g$ on $\partial \mathbb{R}_{+}^{n}$ instead, then the solution is
$$u(x)=\int_{\partial \mathbb{R}_{+}^{n}}K(x,y)g(y)dy$$
where $K(x,y)$ is the Poisson kernel
$$K(x,y)=\frac{2x_{n}}{n\alpha(n)}\frac{1}{|x-y|^{n}}$$
You can read about all of that and more, again, in Evans, Chapter 2.2.4.b (in particular on page 37).
