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There is a expression given in article arXiv:1505.01908 $$\bigg[n\big(\omega+\frac{\Omega}{2}\big)-n\big(\omega-\frac{\Omega}{2}\big)\bigg]g^R_{\omega-\frac{\Omega}{2}}g^A_{\omega+\frac{\Omega}{2}} +n\big(\omega-\frac{\Omega}{2}\big)g^A_{\omega-\frac{\Omega}{2}}g^A_{\omega+\frac{\Omega}{2}} -n\big(\omega+\frac{\Omega}{2}\big)g^R_{\omega-\frac{\Omega}{2}}g^R_{\omega+\frac{\Omega}{2}} \tag{45a} $$ here $n(\omega)=[e^{\beta\omega}-1]^{-1}$, $n'(\omega)=dn/d\omega$, and $$ g_\omega^{R(A)}=\frac{1}{\omega-\omega_q\pm i\alpha\omega} $$ they claim that in limit $\Omega\to 0$, (neglecting $\Omega^2$ terms), the above expression is equal to $$ \approx -\frac{\Omega}{2}n'(\omega)\bigg(g_\omega^A-g_\omega^R\bigg)^2 +n(\omega)\big[(g^A_{\omega})^2-(g^R_{\omega})^2\big] \tag{45b}$$

I wonder how exactly they get the final expression.

My Attempt: I can see that if we take limit $\Omega\to 0$ in the last two terms of $(45a)$, we get the last the last two terms of $(45b)$. And by using approximation for derivative $f'(x)=[f(x+h)-f(x-h)]/2h$, we get $$ \bigg[n\big(\omega+\frac{\Omega}{2}\big)-n\big(\omega-\frac{\Omega}{2}\big)\bigg]=\Omega \frac{dn(\omega)}{d\omega}$$ all in all, so far, I have

$$ \approx \Omega n'(\omega)\lim_{\Omega\to 0}\bigg(g^R_{\omega-\frac{\Omega}{2}}g^A_{\omega+\frac{\Omega}{2}}\bigg) +n(\omega)\big[(g^A_{\omega})^2-(g^R_{\omega})^2\big] \tag{A}$$ I am not sure how is the first term handled to get the expression given in $(45b)$

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    $\begingroup$ Just expand everything to first order as a Taylor series in $\Omega$ and drop terms of order larger than $\Omega$. That will do it. $\endgroup$
    – march
    Jan 10, 2022 at 23:35
  • $\begingroup$ @march thank you so much! I got the answer. Perhaps, the approach that I've explained in my question is not completely correct. The correct way is to apply Taylor expansion of all entities involved. $\endgroup$ Jan 11, 2022 at 14:28
  • $\begingroup$ @march I have come again to thank you. You have no idea how much helpful your one single comment was. You are the real MVP. $\endgroup$ Jan 11, 2022 at 22:00
  • $\begingroup$ The moral of the story is to always remember that if something is small, then you're probably going to use a Taylor series expansion in that small parameter. It's everywhere in physics. It should be right at the top of your toolbox. (Note also that this is essentially equivalent with how you were trying to do it; Taylor series are more mechanical though and so easier to use.) $\endgroup$
    – march
    Jan 11, 2022 at 22:23

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