Large $x$ expansion of $\ln \frac{1+e^{ax}}{1+e^{-bx}}$

Question

When $x$ is very large, the following expression (with $a>0$, $b>0$) $$\ln \frac{1+e^{ax}}{1+e^{-bx}}$$ Can be approximated using $$1+e^{ax}\approx e^{ax}$$ $$1+e^{-bx}\approx 1$$ Therefore $$\ln \frac{1+e^{ax}}{1+e^{-bx}}\approx \ln e^{ax}=ax$$ This is the leading term. How I can get the next term in the approximation? I mean $$\ln \frac{1+e^{ax}}{1+e^{-bx}}\approx ax +c+\frac{c_1}{x}+\dots$$ I want to know $c$.

Answer 1

$$ \lim_{x\to +\infty}\left[\log\left(\frac{1+e^{ax}}{1+e^{-bx}}\right)-ax\right]=\log\lim_{x\to +\infty}\frac{1+e^{-ax}}{1+e^{-bx}}=0 $$ gives $c=0$.

Answer 2

We have that

$$\ln \left(\frac{1+e^{ax}}{1+e^{-bx}}\right)=\ln\left(1+e^{ax}\right)-\ln\left(1+e^{-bx}\right)=$$ $$=\ln e^{ax}+\ln\left(1+e^{-ax}\right)-\ln\left(1+e^{-bx}\right)=$$ $$= ax+e^{-ax}+o(e^{-ax})-e^{-bx}+o(e^{-bx})$$

therefore we have $c=0$ and also $c_1=0$.

Answer 3

$$\ln(1+e^{ax})-\ln(1+e^{-bx)})-ax=\ln(1+e^{-ax})-\ln(1+e^{-bx)})\approx e^{-ax}-e^{-bx}.$$

This function is very quickly decreasing and the coefficients of its Laurent series are null.