Asymptotic analysis of an expression involving a Fox's H function

Question

One of the performance metrics calculated in the analysis of telecommunications systems is the ergodic channel capacity, $C_{\rm erg}$. During one of my studies, I found the expression below for such a metric considering a Rayleigh channel and AWGN.

\begin{align}\label{eq:CAPACIDADEERGODICA} C_{\rm erg} & = \frac{ z^2 }{2 \bar{\gamma} h_{\rm l}^{2}A_{0}^{2}\log(2)} {\rm H}_{3,4}^{4,1}\left[ \frac{1}{\bar{\gamma}h_{\rm l}^{2}A_{0}^2} \ \middle\vert \begin{array}{c} (-1, 1), (\frac{z^2}{2}, 1), (0, 1) \\ (0, 1), (\frac{z^2}{2}-1, 1), (-1, 1), (-1, 1) \end{array}\right] \end{align}

To provide more insights into high SNR regimes ($\bar{\gamma} \to \infty$), the asymptotic behavior of this metric is often required.

Therefore, I'd like to know the asymptotic expression of the ergodic capacity, given by the above equation when $\bar{\gamma} \to \infty$.

OBS.: All the other variables can be considered constant.

UPDATE 10/01/2023

Additional information on how the previous equation is derived.

The ergodic capacity is found as the expectation of $\log_2(1+\gamma)$: \begin{equation}\label{eq:Capacidade} C_{\rm erg} = \int_{0}^{\infty}\log_2(1+\gamma)f_{\Gamma}(\gamma){\rm d}\gamma, \end{equation} where \begin{align}\label{eq:PDFsnr} f_{\Gamma}(\gamma) & = \frac{ z^2}{2 \bar{\gamma} h_{\rm l}^{2}A_{0}^{2}} {\rm H}_{1,2}^{2,0}\left[ \frac{\gamma}{\bar{\gamma} h_{\rm l}^{2}A_{0}^2} \ \middle\vert \begin{array}{c} (\frac{z^2}{2},1) \\ (0, 1), (\frac{z^2}{2}-1,1) \end{array}\right] \end{align} is the PDF of the instantaneous SNR.

Answer 1

Your expression can be simplified as follows: As $m=q=4$, the denominator parameter $(0,1)$ cancels with the numerator parameter $(0,1)$. Furthermore, as all linear coefficients are one, the Fox H function reduces to the simpler Meijer G function. I'll denote $$\tag{1} \xi = \frac 1 {\bar\gamma h_1^2 A_0^2},\quad \zeta = \frac{z^2}{2}, $$ such that \begin{align}\tag{2a} C_{\rm erg} & = \frac{ \zeta \xi}{\log 2} {\rm G}_{2,3}^{3,1}\left[ \xi \, \middle\vert \begin{array}{c} -1, \zeta \\ -1, -1, \zeta-1 \end{array}\right] \\ &= \frac{\zeta}{\log 2} {\rm G}_{2,3}^{3,1}\left[ \xi \, \middle\vert \begin{array}{c} 0, \zeta+1 \\ 0, 0, \zeta \end{array}\right]\tag{2b}\\ &= \frac{1}{\log 2}\left( \frac {\pi \, \Gamma(1-\zeta)} {\sin(\pi\zeta)} \xi^\zeta -\gamma_\mathrm E - \zeta^{-1} - \log \xi + \mathcal O(\xi) \right),\tag{2c} \end{align} with Euler's constant $\gamma_\mathrm E$, the Gamma function $\Gamma$, and with the usual big $\mathcal O$ notation. The calculation was done in Mathematica.