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Let $\varphi\in C_c^\infty(\Bbb{R^2})$ (infinitely differentiable functions with compact support) and consider $$ I=\int_{\Bbb{R}^2}\Delta\varphi(x)\log|x|^2\ dx, $$ the existence of which is given by the fact that $\log |x|^2$ is locally integrable. How far can one simplify this integral using the Green's identity?


Let $R>0$ be large enough so that $\{|x|<R\}$ contains the support of $\varphi$. Denote $\Omega_{\varepsilon}=\{\varepsilon<|x|<R\}$ and $g(x)=\log|x|^2$

Then $$ I=\lim_{\varepsilon\to 0+}\int_{\Omega_\varepsilon}\Delta\varphi(x)\log|x|^2\ dx =\lim_{\varepsilon\to 0+}\left(\int_{\Omega_\varepsilon}\Delta(\log|x|^2)\cdot \varphi(x) dx +\int_{\partial\Omega_{\varepsilon}}g\frac{\partial\varphi}{\partial\nu}-\varphi\frac{\partial g}{\partial\nu}ds\right) $$ where $\frac{\partial\varphi}{\partial\nu}$ is the normal derivative. Note that the first term is zero since $\Delta g=0$ on $\Omega_\varepsilon$. Let $\Gamma_\varepsilon=\{|x|=\varepsilon\}$ and $\Gamma_2=\{|x|=R\}$. Then by straightforward calculations, one has $$ \lim_{\varepsilon\to 0}\int_{\Gamma_\varepsilon}g\frac{\partial\varphi}{\partial\nu}-\varphi\frac{\partial g}{\partial\nu}ds=0. $$ Everything now boils down to the calculation of $$ \lim_{\varepsilon\to 0}\int_{\Gamma_2}g\frac{\partial\varphi}{\partial\nu}-\varphi\frac{\partial g}{\partial\nu}ds =\int_{\Gamma_2}\log R^2\frac{\partial\varphi}{\partial\nu}-\varphi\frac{2}{R}ds. $$ Could anyone simplify it further?

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HINT:

Note that we have

$$\frac{\partial g}{\partial \nu}=\frac{2}{\epsilon}$$

on $\Gamma_{\epsilon}$. Therefore,

$$\lim_{\epsilon\to 0}\int_{\Gamma_{\epsilon}}\left(g\frac{\partial\phi}{\partial\nu}-\phi\frac{\partial g}{\partial\nu}\right)\,ds=-4\pi\phi(0)$$

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If $R$ is large enough $\varphi$ and $\nabla \varphi$ are equal to $0$ on $\Gamma_2$, and so is the RHS in the last formula.

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