- #1

evinda

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\frac{1}{(2-n)w_n}|x-\xi|^{2-n} &, n \geq 3 \\ \\

\frac{1}{2 \pi} \ln{|x-\xi|} &, n=2

\end{matrix}\right.$

So when $ |x- \xi|=\epsilon $ then

$$E(|x- \xi|)=\left\{\begin{matrix}

\frac{1}{(2-n)w_n}\epsilon^{2-n} &, n \geq 3 \\ \\

\frac{1}{2 \pi} \ln{\epsilon} &, n=2

\end{matrix}\right.$$I want to show that $\lim_{\epsilon \to 0} \int_{\partial{B_{\epsilon}}} E(|x- \xi|) f(\xi)ds=0$ for any continuous and bounded $f$.

There is a hint that we should check the integral $\int_{|x|=\epsilon} E(|x|) ds=0$ by using spherical $(n \geq 3)$ or polar $(n=2)$ coordinates.So, suppose that $n \geq 3$.

Then I thought the following:

$\int_{|x|=\epsilon} E(|x|) ds=\int_{|x|=\epsilon} \frac{1}{(2-n) w_n} |x|^{2-n} ds=\frac{1}{(2-n) w_n} \int_{|x|=\epsilon} |x|^{2-n} ds=\frac{1}{(2-n) w_n} \int_{|x|=\epsilon} \epsilon^{2-n} ds=\frac{\epsilon^{2-n}}{(2-n) w_n} 2 \pi \epsilon^{n-1}=\frac{2 \pi \epsilon}{(2-n) w_n} \to 0 \text{ while } \epsilon \to 0$.

But I didn't use spherical coordinates. Have I done something wrong? (Thinking)