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Let $$F(x,y)=(-1/8\pi)|x-y|^2\ln|x-y|, $$ where $x=(x_1,x_2)$ and $y=(y_1,y_2)$. I want to compute this integral

$$I=\int_A \Delta_x F(x,y)dy,\quad\text{ and }\quad J=\int_A F(x,y)dy,$$ where $\Delta_x=\frac{\partial^2}{\partial x_1^2}+\frac{\partial^2}{\partial x_2^2}$ is Laplace operator and $A:=\{(t,0)\times\{0\}, -1<t<1 \}$.


My try is the following: If $x, y\in A$ then $x=(x_1,0)$ and $y=(y_1,0)$. In this case $F(x,y)=(-1/8\pi)(x_1-y_1)\ln|x_1-y_1|$, then $\Delta_x F(x,y)=(-1/8\pi)\frac{1}{x_1-y_1}$ this implis, $$ I=(-1/8\pi)\int_{-1}^1\frac{1}{x_1-y_1}dy_1=...$$ Is it true??

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  • $\begingroup$ Same one have any idea, I'm stack... $\endgroup$ – Achaire Oct 31 '16 at 10:59

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