According to Wikipedia, if you have two independant Gaussian/normal random variables $X \sim N\left(\mu_X=\nu\cos\theta,\sigma_X^2\right)$ and $Y \sim N\left(\mu_Y=\nu \sin\theta,\sigma_Y^2\right)$. With joint distribution: $$f_{X,Y}(x,y) = \frac{1}{2 \pi \sigma_X \sigma_Y } \mathrm{e}^{ -\frac{1}{2}\left[ (\frac{x-\mu_X}{\sigma_X})^2 + (\frac{y-\mu_Y}{\sigma_Y})^2 \right] }$$
If $\sigma=\sigma_x=\sigma_y$, you can get the distribution of $R = \sqrt{X^2 + Y^2}$ with the Rice or Rician distribution:
$$R \sim \mathrm{Rice}\left(|\nu|,\sigma\right)$$
The distribution is $$f_Z(z\mid\nu,\sigma) = \frac{z}{\sigma^2}\mathrm{e}^{\frac{-(z^2+\nu^2)} {2\sigma^2}}I_0\left(\frac{z\nu}{\sigma^2}\right)$$and its mean is: $$\mu_1^{'}= \sigma \sqrt{\pi/2}L_{1/2}(-\nu^2/2\sigma^2)$$
The derivation is given in example 6-16 of this textbook.
My question is the following:
How to derive a compact expression of the mean of the same distribution with $\sigma_x \ne \sigma_y $ ?
I've try to calculate it with Mathematica but failed:
If I substitute the $\sigma_Y=1*\sigma$ by $\sigma_Y=2*\sigma$ the integral cannot be calculated.