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I have a Euclidean geometric problem whereby the triple integral over SO(3) I am trying to solve seems to confuse many CAS softwares (including Maxima and Mathematica). The problem is

$ a = \int^{\sigma_{max}}_{-\sigma_{max}} \int^\pi_{-\pi} \int^{\theta_{max}}_0 \frac{\sin\theta d\theta d\phi }{|x+R(\theta,\phi,\sigma)y|^3} (x+R(\theta,\phi,\sigma)y) \cdot Q \cdot (x+R(\theta,\phi,\sigma)y) d\sigma, $

where $a$ is a real number, $x$ and $y$ are 3D, rank-1 vectors, $R$ is a 3D rotation matrix which rotates the vector $y$, {$\theta$, $\phi$, $\sigma$} is a set of $SO(3)$ rotational parameters, and Q is a 3D, rank-2 traceless, symmetric tensor. The rotation matrix $R(\theta, \phi, \sigma)$ elements in this case would be

$ \begin{array}{rcl} R[0, 0] &=& \hphantom{-}\cos\phi\cos\theta\cos(\sigma-\phi) - \sin\phi\sin(\sigma-\phi), \\ R[0, 1] &=& -\cos\phi\cos\theta\sin(\sigma-\phi) - \sin\phi\cos(\sigma-\phi), \\ R[0, 2] &=& \hphantom{-}\cos\phi\sin\theta, \\ R[1, 0] &=& \hphantom{-}\sin\phi\cos\theta\cos(\sigma-\phi) + \cos\phi\sin(\sigma-\phi), \\ R[1, 1] &=& -\sin\phi\cos\theta\sin(\sigma-\phi) + \cos\phi\cos(\sigma-\phi), \\ R[1, 2] &=& \hphantom{-}\sin\phi\sin\theta, \\ R[2, 0] &=& -\sin\theta\cos(\sigma-\phi), \\ R[2, 1] &=& \hphantom{-}\sin\theta \sin(\sigma-\phi), \\ R[2, 2] &=& \hphantom{-}\cos\theta. \end{array} $

In some cases $\theta_{max}$ is also a function of $\phi$, e.g.

$ \frac{1}{\theta_{max}^2} = \frac{\cos^2\phi}{\theta_x^2} + \frac{\sin^2\phi}{\theta_y^2} $

Would anyone have any advice as to how I should approach such a problem? Is such a triple integral solvable without reverting to numerical integration?

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    Sorry (1), yes, in terms of $x$, $y$, $Q$, $\theta_{max}$, $\sigma_{max}$. (2) The rotation matrix $R(\theta, \phi, \sigma)$ in this case would be \left| \begin{array}{ccc} \cos\phi\cos\theta\cos(\sigma-\phi) - \sin\phi\sin(\sigma-\phi) & -\cos\phi\cos\theta\sin(\sigma-\phi) - \sin\phi\cos(\sigma-\phi) & \cos\phi\sin\theta \\ \sin\phi\cos\theta\cos(\sigma-\phi) + \cos\phi\sin(\sigma-\phi) & -\sin\phi\cos\theta\sin(\sigma-\phi) + \cos\phi\cos(\sigma-\phi) & \sin\phi\sin\theta \\ -\sin\theta\cos(\sigma-\phi) & \sin\theta \sin(\sigma-\phi) & \cos\theta \end{array} \right| 2012-07-26

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