# A simple integral on the hyper-ellipsoid

Let $A \in \mathbb{R}^{n\times n}$ be a positive definite matrix. I want to evaluate the following integral. (This is part of a normalizer for a probability distribution.) $$\int_{||u|| = 1} \frac{1}{\sqrt{u^\top A u}} \mathrm{d}u$$ This is the inverse of $A$ norm integrated over a the unit hypersphere. By a change of variable, $x = A^{1/2} u$, we can rewrite it as, $$|A|^{1/2} \int_{||A^{-1/2} x|| = 1} \frac{1}{\sqrt{x^\top x}} \mathrm{d}x$$ which is an integral on a hyper-ellipsoid. Any hints appreciated!

[EDIT] In 2D, wolframalpha tells me that it's a Complete Elliptic Integral of the First Kind.

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perhaps, switch to n-spherical polar coordinates? –  Valentin Jan 26 '13 at 22:46
@Valentin Thanks for the suggestion. In n-spherical coordinates, the norm becomes simple, but interaction with $A$ is not very friendly, or is it? –  Memming Jan 26 '13 at 23:21

I'm afraid that there might not be a good closed form solution. $\sqrt{u^TAu}$ has lots of branch cuts on the imaginary axis of the $u$ plane, which would be quite difficult to deal with. By using 4.646 of Gradshteyn and Ryzhik, I got something similar to $\sim\text{const}\cdot\int_0^\infty\frac{dx}{\sqrt{\det{(\mathbf{1}+x^2 A})}}$, but I am not sure how useful this could be.

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The first thing I would do is to do an orthogonal transformation:

$A = Q \Lambda Q^{\top}$, with orthogonal $Q$ and diagonal $\Lambda$.
The Q then cancels out and the integral is then $\int_{||x||=1} \frac{1}{\sum_{i=1}^d \lambda_i x_i^2} dx$
Not sure yet if it is nicely integrable thought (should look at $d=2$ and $d=3$ cases to get it)

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