Does anyone know how to handle this integral: $$\int d^3 r \int d^3 r' \phi(\mathbf{r})\phi(\mathbf{r'})\frac{Y^{0}_{2}(\mathbf{r} - \mathbf{r'})}{|\mathbf{r} - \mathbf{r'}|^3}$$ with $$\phi(\mathbf{r}) = \exp\left( -\frac{x^2}{\sigma_{1}^{2}} -\frac{y^2}{\sigma_{2}^{2}} -\frac{z^2}{\sigma_{3}^{2}}\right)$$ and $Y^{2}_0(\mathbf{r})$ being spherical harmonic: $$Y_{2}^{0}(\mathbf{r}) = \frac{1}{4}\sqrt{\frac{5}{\pi}}\frac{2z^2 - x^2 - y^2}{x^2 + y^2 + z^2}$$ All variables extent from $-\infty$ to $+\infty$. I don't really know if any analitical formula exists. Maybe some series expansion of Green's function ? Please help.

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    $\begingroup$ Can you be a little more explicit about the spherical harmonic? For example, I do not understand your notation. Do you mean $$Y_2^0(\theta, \phi) = \frac14 \sqrt{\frac{5}{\pi}} (3 \cos^2{\theta}-1)$$ $Y_0^2$ makes no sense in terms of the notation I understand. $\endgroup$ – Ron Gordon Oct 16 '14 at 23:29
  • $\begingroup$ exactly! I edited the question $\endgroup$ – WoofDoggy Oct 16 '14 at 23:33
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    $\begingroup$ Can you give some context? This reminds me of some computations in quantum mechanics, and if that's the case then it's worth tagging it as such. $\endgroup$ – Semiclassical Oct 16 '14 at 23:49
  • $\begingroup$ Actually it is connected with BEC and dipol-dipol interactions. This $\phi$ is the condensate wavefunction. This is just one of the three integrals. The other two have different harmonics $\endgroup$ – WoofDoggy Oct 17 '14 at 0:09

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