What is the well-known formula? I am reading a paper and I puzzled with the following formula :
Suppose $g\in C^2 (R^2)$ with compact support,show
$$-\frac{1}{2 \pi} \iint_{R^2} \Delta g(z) \log \frac{1}{|z-\xi |}dxdy =g(\xi)$$
I will appreciate for your help.
 A: I don't know the name of this formula but it is the two dimension counterpart for
a well known integral identity in electrostatics.
$$ -\frac{1}{4\pi}\int_{\mathbb{R}^3} \frac{1}{|\vec{x} - \vec{y}|} \vec{\nabla}^2 \phi(\vec{x})\; d\vec{x} =
\phi(\vec{y})$$
The imporant point is $\;\displaystyle \log\frac{1}{|z - \xi|}\;$ is a solution to the two dimension Poisson equation with a point source at $\xi$.
Let $C_r = \{\; z \in \mathbb{R}^2 : |z - \xi| = r\;\}$ 
and $V_r = \{\; z \in \mathbb{R}^2 : |z - \xi| \ge r\;\}$
be the circle centered at $\xi$ with radius $r$ and the region outside it.
Since $\log\frac{1}{|z - \xi|}$ only diverges logarithmically as $z \to \xi$, we have
$$-\frac{1}{2\pi} \int_{R^2} \Delta g(z)\log\frac{1}{|z-\xi|} dxdy
= -\frac{1}{2\pi} \lim_{r\to 0}\int_{V_r} \Delta g(z)\log\frac{1}{|z-\xi|} dxdy
\tag{*1}
$$
Using the fact
$$\Delta \log\frac{1}{|z-\xi|} = 0\quad\text{ for } z \in \mathbb{R}^2 \setminus \{ \xi \},$$ we can replace the integrand in RHS$(*1)$ by
$$ \log\frac{1}{|z-\xi|} \Delta g(z) - 
g(z) \Delta \log\frac{1}{|z-\xi|}
= \vec{\nabla} \cdot \left( 
\log\frac{1}{|z-\xi|} \vec{\nabla} g(z) - g(z) \vec{\nabla} \log\frac{1}{|z-\xi|}
\right)$$
Apply the two dimension version of 
divergence theorem and using
the fact $g(z)$ has compact support, the integral in RHS of $(*1)$ becomes
$$\begin{align}
& -\frac{1}{2\pi} \left( \int_{C_\infty} - \int_{C_r} \right)
\left( 
\log\frac{1}{|z-\xi|} \vec{\nabla} g(z) - g(z) \vec{\nabla} \log\frac{1}{|z-\xi|}
\right) \cdot \vec{n} dS\\
= &
\frac{1}{2\pi} \int_{C_r}
\left( 
\log\frac{1}{|z-\xi|} \vec{\nabla} g(z) - g(z) \vec{\nabla} \log\frac{1}{|z-\xi|}
\right) \cdot \vec{n} dS\\
= & \frac{1}{2\pi} \left[
\log\frac{1}{r}\left(\int_{C_r} \vec{n}\cdot\vec{\nabla} g(z) dS\right)
+ \frac{1}{r}\left(\int_{C_r} g(z) dS\right)\right] \\
\end{align}\tag{*2}
$$
where $\vec{n}$ and $dS$ are the outward pointing normals and line element for the circles $C_{\infty}$ and $C_{r}$.
Inside the square bracket of $(*2)$, we know


*

*the $1^{st}$ integral behaves like $O(r)$ as $r \to 0$.

*the $2^{nd}$ integral behaves like $2\pi g(\xi)r + O(r^2)$ as $r \to 0$.


Combine these, we can evaluate the RHS of $(*1)$ as
$$\lim_{r\to 0} \frac{1}{2\pi} \left[ O(r\log\frac{1}{r}) + 2\pi g(\xi) + O(r) \right] = g(\xi)$$
A: I just want to add that as achille hui suggested, $$u=-\frac{\log(r)}{2\pi}$$is the fundamental solution of Laplace's equation on $\mathbb{R}^{2}$. So your formula is just
$$
\int u*\triangle g=\int \triangle (u*g)=\int (\triangle u)*g=g(\xi)
$$
where the origin shifted to $\xi$. If I made a mistake please point out, as then I can correct it. 
I think the derivation of this formula under radial symmetry is one of the basic examples in Evan's book (Chapter 2). So maybe you can look it up. 
