Let $A$ and $b$ be an $M\times M$ matrix and an $M\times 1$ vector, respectively.
I need to solve $$\int_{\|x\|^2=1} \exp\left(-x^\ast Ax + 2\mathcal{R}\{x^\ast b\}\right) \, dx$$ where $(\cdot )^\ast$ is conjugate transpose of a vector and $\mathcal{R}\{ \cdot\}$ takes the real part.
Is this doable. Seems related to the Itzykson-zuber integral, but I cannot find anything in the literature.
I doubt that there is a closed-form expression for any $M$, but for large $M$ you can approximate the integral by replacing the constraint $\|x\|=1$ by a Gaussian measure
uncorrelated $x_n$'s with mean and variance $E(x_n)=0$, $E(|x_n|^2)=1/M$
I assume that $A$ is positive definite, with eigenvalues $a_n>0$, $n=1,2,\ldots M$. Note that the integral depends only on the eigenvalues of $A$, not on the eigenvectors, so we may as well take $A$ diagonal. The vector $b=(b_1,b_2,\ldots b_M)$ is then taken in that basis. It is convenient to normalize the integral by the area $S_M$ of the $M$-dimensional unit hypersphere.
I then arrive at
$$\frac{1}{S_M}\int_{\|x\|^2=1} \exp\left(-x^\ast Ax + 2\mathcal{R}\{x^\ast b\}\right)dx$$
$$=\frac{e^{b^\ast A^{-1}b}}{S_M}\int_{\|x\|^2=1}\exp[-(x-A^{-1}b)^\ast A(x-A^{-1}b)]\,dx\rightarrow$$
$$\rightarrow\prod_{n=1}^M\left(e^{|b_n|^2/a_n}\int_{-\infty}^\infty\exp[-(x-b_n/a_n)^\ast a_n(x-b_n/a_n)]e^{-|x|^2/M}\,\frac{dx}{\pi M}\right)$$
$$=\prod_{n=1}^M\frac{\exp\left(\frac{M|b_n|^2}{1+Ma_n}\right)}{1+Ma_n}$$
