Bob Oakley

I try to simplify the following singular integral:

>$$\int_{R^2}\hat{p}(\xi)\hat{q}(\xi)|\xi|^{-2}d\xi= A\int_{R^2}\int_{R^2}p(x) q(y) (-\log|x-y|)dxdy$$
where $p, q\in C_c^{\infty}$ and $\int q=0$ and $\hat{p}(\xi):= (1/(2\pi))\int e^{-ix\xi}p(x)dx$ which is the Fourier transform, and $A$ is a constant.

1. My way is LHS=
$$\int_{R^2\setminus B(\epsilon)}\hat{p}(\xi)\hat{q}(\xi)|\xi|^{-2}d\xi+\int_{B(\epsilon)}\hat{p}(\xi)\hat{q}(\xi)|\xi|^{-2}d\xi$$
But why the RHS appears $(-\log|x-y|)$ which is the log potential of Poisson equation.

I try to simplify the following singular integral:

>$$\int_{R^2}\hat{p}(\xi)\hat{q}(\xi)|\xi|^{-2}d\xi= A\int_{R^2}\int_{R^2}p(x) q(y) (-\log|x-y|)dxdy$$
where $p, q\in C_c^{\infty}$ and $\int q=0$ and $\hat{p}(\xi):= (1/(2\pi))\int e^{-ix\xi}p(x)dx$ which is the Fourier transform, and $A$ is a constant.

1. My way is LHS=
$$\int_{R^2\setminus B(\epsilon)}\hat{p}(\xi)\hat{q}(\xi)|\xi|^{-2}d\xi+\int_{B(\epsilon)}\hat{p}(\xi)\hat{q}(\xi)|\xi|^{-2}d\xi$$
But why the RHS appears $(-\log|x-y|)$ which is the log potential of Poisson equation.

Do you think if it is right?
Here is the key point: LIM((\frac{x_1-x_{n+1}}{n})+B(x_2,x_3,x_4,\dots)=B(x_2,x_3,x_4,\dots) since $\Vert x_1-x_{n+1}\Vert \leq 2\Vert x\Vert_{\infty}$ which is bounded. Then LIM (\frac{x_1-x_{n+1}}{n})=0
Then I can show that $B(\{x_{n+1}\})=B(\{x_{n}\})$.
Remark: I use B(\{x_{n}\})=B(x_1,x_2,x_3,\dots)=B(x_1-x_2, x_2-x_3,x_3-x_4,\dots)+B(x_2,x_3,x_4,\dots)=LIM(\frac{1}{n}\sum_{i=1}^n(x_i-x_{i+1}))+B(x_2,x_3,x_4,\dots)
Denote the shift operator $T: l_{\infty}\to l_{\infty}$ by \[T: (x_1, x_2, x_3, \dots)\to (x_2, x_3, x_4, \dots).\] So we have $B(T\{x_{n}\})=B(\…