Finding the General Solution to the Dirichlet Problem for the Poisson Equation I am trying to show, with the following PDE:
$$\begin{cases}\Delta u=f &\text{on}\>\>\Omega\\
u=g &\text{ on }\partial\Omega
\end{cases}$$
that for all $x_0\in\Omega$, we have:
$$u(x_0) = \iiint_{\Omega}G(x,x_0)f(x) dx+\iint_{\partial\Omega}g(x)\frac{\partial}{\partial \vec n}G(x,x_0)dS$$
Where $\vec n$ is the normal vector to $\partial\Omega$. 
Now, we have defined a Green's Function to be:
$$G(x,x_0) = H(x)+\Phi(x)$$
Where $\Phi(x)$ is the Fundamental Solution in the respective Dimension. We also have the characteristics of $G$ and $H$, $H$ is a smooth Harmonic function in all of $\Omega$, and $G(x,x_0)=0$ on $\partial\Omega$. 
For simplicity let us assume we are in Dimension $3$. Then, I believe we can solve this by Superposition. We had previously solved:
$$\begin{cases}\Delta u=0 &\text{on}\>\>\Omega\\
u=g &\text{ on }\partial\Omega
\end{cases}$$
The solution was given by:
$$u(x_0)=\iint_{\partial\Omega}g(x)\frac{\partial}{\partial\vec n}G(x,x_0)dS$$
Good. This is half of what we need to show. Now, by Superposition I need to find the answer to:
$$\begin{cases}\Delta u=f &\text{on}\>\>\Omega\\
u=0 &\text{ on }\partial\Omega
\end{cases}$$
I started by applying Green's Second Identity to $H(x)$ and $u$ on $\Omega$:
$$\iiint_{\Omega}H\Delta u-\Delta H u dx = \iint_{\partial\Omega}H\frac{\partial u}{\partial\vec n} dS_x-\iint_{\partial\Omega}\frac{\partial H}{\partial \vec n} u dS_x$$
Now, we know that $\Delta H$ is zero on $\Omega$. I also know that $u$ and all its derivatives are $0$ on $\partial\Omega$ by Superposition. So I am left with:
$$\iiint_{\Omega}H\Delta u dx=\iiint_{\Omega}G\Delta u dx -\iiint_{\Omega}\Phi(x) \Delta u dx$$
Now I know above that $\Delta u = f$. All I need to show is that the above second term on the right hand side goes to $0$. But how do I show this??
 A: You say that you know that $u$ is zero on the boundary $\partial \Omega$ and that you want to show that:
$$ \int_\Omega \Phi(x) \Delta u \ \mathrm{d}x = 0$$
This is just a classic application of Green's billionth identity. We have:
$$ \int_\Omega \Phi(x) \Delta u \ \mathrm{d} x = \int_\Omega u(x) \Delta \Phi \ \mathrm{d} x+ \int_{\partial\Omega} \Phi (\nabla u \cdot \mathbf{\hat{n}}) - u (\nabla \Phi \cdot \mathbf{\hat{n}}) \ \mathrm{d} S$$
Since $\Delta \Phi = 0$ in $\Omega$ and $u = 0$ on $\partial \Omega$, this leaves us with the following integral:
$$ \int_{\partial\Omega} \Phi (\nabla u \cdot \mathbf{\hat{n}}) \ \mathrm{d} S $$
Note that we have the following identity by the product rule:
$$ \nabla(\Phi u) = \Phi \nabla u + u \nabla \Phi $$
So that $\Phi \nabla u = \nabla(\Phi u ) - u\nabla \Phi $. If we now plug this into our integral we get:
$$ \int_{\partial\Omega} \Phi (\nabla u \cdot \mathbf{\hat{n}}) \ \mathrm{d} S = \int_{\partial\Omega} \nabla(\Phi u ) \cdot \mathbf{\hat{n}} \ \mathrm{d} S - \int_{\partial\Omega} u \nabla\Phi  \cdot \mathbf{\hat{n}} \ \mathrm{d} S = 0$$
The first integral vanishes, since we are integrating the conservative vector field $\nabla (\Phi u)$ over a closed curve $\partial\Omega$, and the second integral vanishes since $u = 0$ on $\partial \Omega$. The conclusion is then:
$$ \int_\Omega \Phi(x) \Delta u \ \mathrm{d}x = 0$$
There you go! This is exactly what you wanted to show.
A: I strongly advise you have a close look at the following lecture notes taylor made for Green's  functions. It will certainly solve your problem. I fact having read this I realized you did not take care of singularities occurred by the fundamental solution that is why you problem to show that the term $$\iiint_{\Omega}G\Delta u dx -\iiint_{\Omega}\Phi(x) \Delta u dx$$ cancel out.  All details about that issues is given therein 
