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Let $x\in \mathbb R^n$ and $B_\varepsilon = \{x:\|x\|\leq \varepsilon\}$. In the proof of the solution for a Poisson equation $$ -\Delta u = f $$ there is written that $$ \int\limits_{B_\varepsilon}|\Phi(y)|dy\leq C\varepsilon^2 $$ for $n\geq 3$. Here $\Phi(y)$ is a fundamental solution of Laplace equation. Since $\Phi(y) = k |y|^{2-n}$ I used to think that for $n>4$ the integral will diverge since $\int\limits_{0}^\varepsilon y^{-2}dy = \infty$. Do I have the wrong reasoning?

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up vote 5 down vote accepted

The fundamental solution of the Laplace equation is $|y|^{2-n}$ (exponent $2-n$, not $n-2$). In your last integral, you're missing the factor from the Jacobian determinant. The correct integral is

$$\iint_0^\epsilon|y|^{2-n}|y|^{n-1}\mathrm d|y|\mathrm d\Omega=\iint_0^\epsilon|y|\mathrm d|y|\mathrm d\Omega= \frac12\epsilon^2\int\mathrm d\Omega=C\epsilon^2\;.$$

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Thanks. $n-2$ was a misprint –  Ilya Aug 1 '11 at 15:12
    
Does it informally mean that it higher dimensions the area of $B_\varepsilon$ decreases faster then $y^{2-n}$ grows? –  Ilya Aug 1 '11 at 15:22
    
If by "area" you mean "surface area", yes, the surface area of $B_\epsilon$ is proportional to $\epsilon^{n-1}$, which is one factor of $\epsilon$ more in the numerator than $y^{2-n}$ has in the denominator. –  joriki Aug 1 '11 at 15:28

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