I want to prove that $$ \mathbb{E} \Big\{ \exp\left(-x\lVert{\mathbf{H}}\rVert_F^2\right)\Big\} = \frac{1}{\det(\mathbf{I}_{m,n}+x\mathbf{R})}=\prod_{i=1}^{m\times n}\frac{1}{1+x\lambda_i(\mathbf{R})}, $$ where $\mathbf{H}$ is $m\times n$ complex-valued matrix whose each element obeys Gaussian distribution with zero-mean and unit variance, $\lVert\cdot\rVert_F$ stands for the Frobenius norm, so $\lVert\mathbf{H}\rVert_F^2=\sum_{i=1}^m\sum_{j=1}^n\lvert h_{i, j}\rvert^2$, $\mathbf{I}_{m,n}$ is the identity matrix with size of $m\times n$, $\mathbf{R}=\mathbb{E}\{\mathrm{vec}(\mathbf{H})\mathrm{vec}(\mathbf{H}^H)\}$, $\mathbf{H}^H$ is the hermitian matrix of $\mathbf{H}$, and $\lambda_i(\mathbf{R})$ implies the $i$th eigenvalue of $\mathbf{R}$.
Is there someone to prove the above theorem?
It's false for $m=n=1$. When $x\not= 0$, $E(\exp(-xh^2))\not= \dfrac{1}{1+xE(h^2)}.$