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In my research, I need to evaluate an integral: $$\int_{R^{3}}\frac{y^{3}}{(|x-\xi|^{2}+y^{2})^{3}}\log(|\xi^{2}|+\frac{1}{4})d\xi$$ where $x\in R^{3}$, $y\geq0$. Moreover, I want to see whether it is a biharmonic function on the up-half space or not. I also want to know will $$ \log(|x|^{2}+(y+\frac{1}{2})^{2})-c\int_{R^{3}} \frac{y^{3}}{(|x-\xi|^{2}+y^{2})^{3}} \log(|\xi^{2}|+\frac{1}{4})d\xi$$ be bounded on the up half space? Here $c$ is a constant such that $$c\int_{R^{3}}\frac{y^{3}}{(|x-\xi|^{2}+y^{2})^{3}}d\xi=1$$

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1 Answer 1

If you use spherical coordinates, the integral splits in a radial part and an integral over the unit sphere in $\mathbb{R}^3$. The spherical integral is $$ \int_{S^2} \frac{1}{y^2+r^2+|x|^2 - 2 r \langle x , \xi' \rangle} d \sigma(\xi') $$where $\xi= r \xi'$, $\xi' \in S^2$. Now you can apply the Funk-Hecke theorem to compute this spherical integral, because the integrand only depends on the inner product of $\xi'$ with a fixed vector.

(you can find the Funk-Hecke theorem in books on spherical harmonics).

| haven't carried out this computation explicitly, but this is how I would compute this integral.

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Thanks, although I still can not evalute that integral, at least the Funk-Hecke theorem can reduce it to 1-d integral. –  wrwrnm Feb 25 '11 at 20:32
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