MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $$c\int_{R^{3}}\frac{y^{3}}{(|x-\xi|^{2}+y^{2})^{3}}d\xi=1$$

share|cite|improve this question

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.

share|cite|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.