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I study an article, and I got stuck on a problem of calculating an integral. Whatever I do, I do not get the result mentioned there. The notations are $u,\tilde u$ are functions defined on $\Omega \subset \Bbb{R}^N$ with values in $\Bbb{R}^n$: $$ \eta_\varepsilon = \int_\Omega \tilde u(x)dx -\int_\Omega u(x)dx \in \Bbb{R}^n $$

The idea is that we try to correct $\tilde u$ such that it has the same integral as $u$. So pick a ball $B_\varepsilon=B(x_0,\varepsilon^{1/N})$ contained in the interior of $\Omega$. We define the function $$ v(x)=\begin{cases} \tilde u(x) & x \notin B_\varepsilon \\ \tilde u(x)+h_\varepsilon(1-\varepsilon^{-1/N}|x-x_0|) & x \in B_\varepsilon \end{cases}$$ where $h_\varepsilon \in \Bbb{R}^n$ should be chosen such that $\int_\Omega v=\int_\Omega u$, i.e. $$ \int_{B_\varepsilon} h_\varepsilon(1-\varepsilon^{-1/N}|x-x_0|)dx=-\eta_\varepsilon $$ Since $h_\varepsilon$ is a constant, it should be enough to find the value of $$(I) \ \ \ \ \int_{B_\varepsilon} (1-\varepsilon^{-1/N}|x-x_0|)dx $$ which I calculated with the coarea formula and got $$ |B_\varepsilon|-\varepsilon^{-1/N}\int_0^{\varepsilon^{1/N}}r^N \cdot N \omega_N dr=\frac{ \varepsilon\omega_N}{N+1}$$ where $\omega_N$ is the volume of the unit ball in $\Bbb{R}^N$. This would lead to $$ h_\varepsilon=-\eta_\varepsilon \frac{N+1}{\varepsilon \omega_N}$$

However, in the article the answer is $$ h_\varepsilon = -\eta_\varepsilon \frac{N}{ \varepsilon^{\frac{N-1}{N}}\omega_{N-1}}$$ This seems wrong to me. The different constants are not a problem but some facts presented in the following of this rely crucially on the fact that the power of $\varepsilon$ in $h_\varepsilon$ is $\frac{1-N}{N}$ and not $-1$ as I got.

Is my calculation of the integral $(I)$ correct, or I missed something?

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up vote 2 down vote accepted

Independent of the details of your calculation, the book's answer can't be right since $|B_\epsilon|$ clearly goes as $\epsilon\omega_N$ and not as $\epsilon^{(N-1)/N}\omega_{N-1}$. It looks as if they were calculating the integral over the sphere rather than the ball, but I don't see why they would do that. Are you sure that $\Omega\subset\mathbb R^n$? Sometimes $\Omega$ is used to denote solid angle.

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The article is available for free here:… The calculation I mentioned is at pages 79-80. $\Omega$ is surely an open subset of $\Bbb{R}^N$ – Beni Bogosel Apr 5 '12 at 15:09
As you say, it can't be the way the author of the article wrote. Probably I'll change the ball to a cylinder of fixed height to get the exponent I need for $\varepsilon$. – Beni Bogosel Apr 5 '12 at 16:54

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