I am going through the calculation by Rawson et al. [J. Opt. Soc. AM. Vol. 70, No. 8, August 1980] and ran into seemingly simple issue with the derivation. I wanted to get some help on solving the problem. The paper calculates intensity profile of K different plane waves propagating through a waveguide, where the total intensity of the light at some distance is given by,
$I(x,\nu) = \sum_{k=-K}^{k=K} \sum_{m=-K}^{m=K} \alpha_k\alpha_m \times e^{i a \nu [b \lambda^2 (k^2-m^2)-c \lambda (k-m)x]}\times e^{i(\phi_k-\phi_m)}$
where $\phi_k = g \times \delta L_k(x)$ represents a random change of the path length ($\delta L_k$) suffered by the kth mode.
We would like to take spatial averaging of the intensity profile ($<f(x)>= 1/2w \int _{-w}^w f(x) dx$) and ensemble average ($\bar{I}$) of the phase variation to get the average intensity profile.
Now the paper assumes that the phases $\phi_k$ and $\phi_m$ are distributed uniformly from ($-\pi, \pi$) and are independent when $k\neq m$. With this assumption they derive,
$<\bar{I(x,\nu)}> = \sum_{k=-K}^{k=K} |\alpha_k|^2 $
Instead of assuming that the phase is uniformly distributed between [$-\pi, \pi$], I would like to calculate the intensity profile assuming that $\delta L_k$ is pulled from a Gaussian distribution with width $\sigma_l$. How would the above integral change?
Any help would be appreciated. Thanks