If the period is much bigger than the half-life, ie. the degradation rate is much bigger than the frequency : $latex \displaystyle{ \Large \gamma \gg \Omega }$ ie. $latex \displaystyle{ \Large \delta << 1 }$, we can make some simplifications.

Let us recall the transcendental equation for the frequency :

$latex \displaystyle{ \Large \Omega = \omega +\frac{S}{2(\delta^2+1)^{\frac{1}{4}}} Im \left( e^{\frac{i}{2} \left( \arctan(\frac{\omega}{\gamma} )-\arctan(\delta) \right)}\left( \frac{1}{1-\exp(-\frac{(\delta^2+1)^{\frac{1}{4}}}{\lambda}e^{\frac{i}{2}\arctan(\delta)} d+i\beta)}+\frac{1}{1-\exp(-\frac{(\delta^2+1)^{\frac{1}{4}}}{\lambda}e^{\frac{i}{2}\arctan(\delta)} d-i\beta)} -1 \right) \right) }$

We do not know if $latex \displaystyle{\Large \frac{\Omega}{\gamma} }$ is negligible in comparison to $latex \displaystyle{\Large \frac{\omega}{\gamma} }$ so we will keep this term and write  $latex \displaystyle{\Large e^{i(\arctan(\frac{\omega}{\gamma})-\arctan(\frac{\Omega}{\gamma}))} = A+iB}$ 

If $latex \displaystyle{ \Large \Omega = \omega+\epsilon, \epsilon << 1 }$ then $latex \displaystyle{A=1 }$ and $latex \displaystyle{B=\epsilon }$ In a simulation, we find that $latex \displaystyle{ \Large \Omega - \omega \sim 10^{-3} }$ for example.

It is useful to define the three following independant parameters :

$latex \displaystyle{\Large \lambda = \sqrt{\frac{D}{\gamma}} }$ (coupling range in meters)

$latex \displaystyle{\Large S=\frac{\alpha}{\sqrt{D\gamma}} }$ (coupling strength in 1/seconds)

At first order in $latex \displaystyle{ \Large \delta }$ :

$latex \displaystyle{\Large \Omega=\omega+\frac{S}{2} Im \left((A+iB)(\frac{1}{1-e^{-(\frac{1+i\delta}{\lambda}+i\beta)}}+\frac{1}{1-e^{-(\frac{1+i\delta}{\lambda}-i\beta)}}-1)\right) = \omega+\frac{S}{2} Im \left((A+iB) \left(\frac{1}{1-e^{-(\frac{1}{\lambda}+i\beta)}}+\frac{1}{1-e^{-(\frac{1}{\lambda}-i\beta)}}-1-i\frac{\delta}{2\lambda} \left( \frac{1}{e^{(\frac{1}{\lambda}-i\beta)}-1}+\frac{1}{e^{(\frac{1}{\lambda}+i\beta)}-1} \right) \right)\right) }$

A common choice of parameters (alpha=50; omega=10; gamma=100.; D=1)would give $latex \displaystyle{\Large \delta \sim 0.1}$ and $latex \displaystyle{\Large B \sim 0.001}$, so we will choose to take the zeroth order in B (consequently, $latex \displaystyle{ \Large A=1 }$)and the first order in $latex \displaystyle{\Large \delta }$.  Since the sum of two complex conjugates is real, taking the imaginary part leads to :

$latex \displaystyle{\Large \Omega=\omega-\frac{S}{4} \frac{\delta}{\lambda}2 Re\left(\frac{1}{e^{(\frac{1}{\lambda}+i\beta)}-1}\right)}$

$latex \displaystyle{\Large \left( \frac{1}{e^{(\frac{1}{\lambda}+i\beta)}-1}\right)= \frac{e^{\frac{1}{\lambda}}\cos(\beta)-1-ie^{\frac{1}{\lambda}}\sin(\beta)}{|e^{(\frac{1}{\lambda}+i\beta)}-1|^2} }$ so $latex \displaystyle{\Large \Omega=\omega-S \frac{\delta}{2\lambda} \frac{e^{\frac{1}{\lambda}}\cos(\beta)-1}{|e^{(\frac{1}{\lambda}+i\beta)}-1|^2}}$

Considering that $latex \displaystyle{\Large \lambda = \sqrt{\frac{D}{\gamma}} }$, in many cases where $latex \displaystyle{\Large \gamma} $ is big, we have $latex \displaystyle{\Large \lambda }$ small, and we can approximate the previous expression taking the leading order in $latex \displaystyle{\Large e^{\frac{1}{\lambda}}} $ when $latex \displaystyle{\Large \beta }$ is not far from 0:

$latex \displaystyle{\Large \Omega=\omega-S \frac{\delta}{2\lambda} \frac{e^{\frac{1}{\lambda}}\cos(\beta)-1}{1-2e^{\frac{1}{\lambda}}\cos(\beta)+e^{\frac{2}{\lambda}} } \sim \omega-S \frac{\delta}{2\lambda} e^{\frac{-1}{\lambda}}\cos(\beta) } $

We can finally solve the equation for $latex \displaystyle{\Large \Omega }$, recalling that $latex \displaystyle{\Large \delta= \frac{\Omega}{\gamma} }$ :

1. The variation of  $latex \displaystyle{\Large \Omega }$ around the frequency $latex \displaystyle{\Large \omega }$ is not too big (which translates to $latex \displaystyle{\Large \alpha }$ not being to great).
2. The degradation rate $latex \displaystyle{\Large \gamma }$ is much bigger than the frequency $latex \displaystyle{\Large \Omega }$
3. The range of the coupling is small ($latex \displaystyle{\Large D/d^2 \ll \gamma}$ (NB: this condition is less stringent than the previous one, because the approximation is much smaller)