Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi, I need to integrate a function on an n-dimensional sphere surface. One way is to use the triangle function like: http://en.wikipedia.org/wiki/N-sphere#Spherical_volume_element, however, it is too complex, do you have any idea how to solve it?

My problem is $\int_{\|{\bf U}\| = 1} \sqrt{ {\bf U}^T {\bf M} {\bf U} } d u_1 \ldots d u_n$, where ${\bf M}$ is a positive semidefinite matrix.

share|improve this question
1  
You can assume that $\bf{M}$ is diagonal, because orthogonal group acts on a sphere in a measure-preserving way, but that doesn't help, because even in two-dimensional case you end up with an elliptic integral. –  Mateusz Wasilewski Jul 22 '12 at 9:33
    
You may get a better response at e.g. math.stackexchange.com, as MO is for questions of research interest (see the FAQ for more suggestions). –  David Roberts Jul 22 '12 at 10:44
    
It's still a nontrivial problem to integrate this numerically. If I did this right it comes down to an elementary multiple of the hyperelliptic integral $$ \int_0^\infty \left[ 1 - \det(1+{\bf M}t^2)^{-1/2} \right] \frac{dt}{t^2}. $$ By the way, the notation $du_1 \cdots du_n$ for the integration element is misleading: it suggests an $n$-dimensional integral, but $\|{\bf U}\|^2 = 1$ is an $n-1$-dimensional manifold. –  Noam D. Elkies Jul 22 '12 at 15:21
    
One possible way is to use Monte Carlo method. –  Andrew Jul 22 '12 at 15:25

1 Answer 1

An analytic expression is given in my paper http://arxiv.org/pdf/math/0403375.pdf (see page 6) in terms the Lauricella hypergeometric function. Note that the $n$-dimensional integral is hopeless to evaluate numerically by Monte-Carlo method ("the curse of dimensionality"), whereas the hypergeometric function is, at the most primitive, expressed as a one-dimensional integral, so numerical evaluation is very fast.

share|improve this answer

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.