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There is a Weyl integration formula that deals with this problem. $\newcommand{\bR}{\mathbb{R}}$ Denote by $G_n$ the space of symmetric $n\times n$ real matrices. It states that for any $O(n)$-invariant $h:G_n\to \bR$ we have

$$(2\pi)^{-m(m+1)/4} 2\pi)^{-n(n+1)/4} \int_{G_n} h(A) e^{-({\rm tr} A^2)/4} dA$$

$$=\frac{1}{Z_n}\int_{\mathbb{R}^n} h(\lambda_1,\dotsc, \lambda_n) e^{-\frac{1}{4} \sum_{j=1}^n \lambda_j^2} {\prod} |\lambda_j-\lambda_k| d\lambda_1\cdots d\lambda_n$$

where the product is taken over $j< k$ and

$$Z_n =2^{\frac{n}{2}}n! \prod_{j=1}^n \Gamma(\frac{j}{2}).$$

(Sorry for the typesetting clumsiness. There seems to be problem with MathJax.)

Now use this formula for

$$h(A)= f(g(A)) e^{\frac{{\rm tr} A^2}{4}}.$$

For more details see Appendix B of this paper and the references therein.

3 added 15 characters in body

There is a Weyl integration formula that deals with this problem. $\newcommand{\bR}{\mathbb{R}}$ Denote by $G_n$ the space of symmetric $n\times n$ real matrices. It states that for any $O(n)$-invariant $h:G_n\to \bR$ we have

$$(2\pi)^{-m(m+1)/4} \int_{G_n} h(A) e^{-({\rm tr} A^2)/4} dA$$

$$=\frac{1}{Z_n}\int_{\mathbb{R}^n} h(\lambda_1,\dotsc, \lambda_n) e^{-\frac{1}{4} \sum_{j=1}^n \lambda_j^2} {\prod} |\lambda_j-\lambda_k| d\lambda_1\cdots d\lambda_n$$

where the product is taken over $j< k$ and

$$Z_n =2^{\frac{n}{2}}n! \prod_{j=1}^n \Gamma(\frac{j}{2}).$$

(Sorry for the typesetting clumsiness. There seems to be problem with MathJax.)

Now use this formula for

$$h(A)= f(g(A)) e^{\frac{{\rm tr} A^2}{4}}.$$

For more details see Appendix B of this paper and the references therein.

2 added 210 characters in body; deleted 9 characters in body

There is a Weyl integration formula that deals with this problem. $\newcommand{\eS}{\mathscr{S}}$ $\newcommand{\bR}{\mathbb{R}}$ Denote by $G_n$ the space of symmetric $n\times n$ real matrices. It states that for any $O(n)$-invariant $h:G_n\to \bR$ we have

$$\frac{1}{ (2\pi)^{ \frac{m(m+1)}{4} } } 2\pi)^{-m(m+1)/4} \int_{G_n} h(A) e^{-\frac{tr A^2}{4} e^{-({\rm tr} dA=\frac{1}{Z_n}\int_{\bR^nA^2)/4} dA$$

$$=\frac{1}{Z_n}\int_{\mathbb{R}^n} h(\lambda_1,\dotsc, \lambda_n) e^{-\frac{1}{4}\sum_{j=1}^n \lambda_j^2 e^{-\frac{1}{4} \prod_{j sum_{j=1}^n \lambda_j^2} {\prod} |\lambda_j-\lambda_k| d\lambda_1\cdots d\lambda_n$$

where the product is taken over $j< k$ and

$$Z_n =2^{\frac{n}{2}}n! \prod_{j=1}^n \Gamma(\frac{j}{2}).$$

Now let

(Sorry for the typesetting clumsiness. There seems to be problem with MathJax.)

Now use this formula for

$$h(A)= f(g(A)) e^{\frac{tr e^{\frac{{\rm tr} A^2}{4}}.$$

For more details see this paper and the references therein.

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