# Can one recognize this symmetric function?

$\newcommand{\lm}{\lambda}$ $\newcommand{\bR}{\mathbb{R}}$ Let $m$ be an integer $>1$. Define

$$I_m:\bR^m\to \bR,\;\; I_m(\lm_1,\dotsc, \lm_m)=\int_{S^{m-1}}\exp\Bigl(-\sum_{j=1}^m \lm_j^2x_j^2\;\Bigr)\; |dA(x)|,$$

where $S^{m-1}$ is the unit sphere in $\bR^m$ and $|dA(x)|$ denotes the "area" element on $S^{m-1}$.

The function $I_m$ is real analytic and symmetric in the variables $\lm_1^2,\dotsc, \lm_m^2$ and in fact it has a Taylor expansion

$$I_m (\lm_1,\dotsc, \lm_m)=2\sum_{h=0}^\infty\frac{(-1)^h}{\Gamma(\frac{m}{2}+h)}\sum_{h_1+\cdots+h_m=h}\frac{\Gamma(h_1+\frac{1}{2})\cdots \Gamma(h_m+\frac{1}{2})}{h_1!\cdots h_m!} \lm_1^{2h_1}\cdots \lm_m^{2h_m}.$$

In particular, $I_m$ can be expressed as a function of the symmetric polynomials

$$s_k =\sum_{j=1}^m \lm_j^{2k},\;\; k=1,\dotsc, m.$$

Question 1. Is there a more compact description of $I_m$ of the form

$$I_m(\lm_1,\dotsc, \lm_m)=F_m(s_1,\dotsc, s_m),$$

where $F_m$ is some "classical" function?

Question 2. $\DeclareMathOperator{\diag}{Diag}$ $\DeclareMathOperator{\tr}{tr}$ Consider the symmetric matrix

$$\Lambda=\diag(\lm_1,\dotsc, \lm_m).$$

Is there some function $V_m:\bR\to \bR$ such that

$$I_m(\lm_1,\dotsc, \lm_m)=e^{-\tr V_m(\Lambda)} ?$$

Can one describe such a $V_m$ explicitly? I'm vague about the term explicit, but I would be very pleased if $V_m$ were a "special" function.

The second question may suggest the origin of $I_m$. I stumbled onto $I_m$ when I bumped into a certain ensemble of random real, symmetric $m\times m$ matrices. (The story is too bushy to include it here.)

Your function is a special case of the Confluent Hypergeometric function of matrix argument, in particular it is ${}_1F_1(\frac{1}{2},\frac{n}{2},L)$, where $L$ is a diagonal matrix having $-\lambda_i^2$ ($1\le i \le n$) as its diagonal components. These functions can be more generally represented using Zonal Polynomials.
It is sufficient to consider the functions $I_m (c,0,\ldots,0)$ since an appropriate rotation (change of variable), where $c$ satisfies $\lambda_1^2 + \cdots + \lambda_m^2 = c^2$, leaves the integral invariant. Things simplify a lot after that.
• Marc, I think there is a problem with your answer: the quantity $\sum_j\lambda_j^2 x_j^2$ is not rotation invariant. Aug 17, 2012 at 19:35
• Liviu, this quantity itself is not rotation invariant, but the integral is invariant since one is integrating over $m-1$-sphere. Aug 17, 2012 at 19:49