Any simplification of $\log_e{(1+e^x)}?$ Is there any simplification or other interesting transformation of:
$$\log_e{(1+e^x)}$$
(where $x \in \mathbb{R}$) ?
 A: You can use the Taylor series of $\log$:
$$ \log(1+e^x) = x+\log(1+e^{-x}) = x + e^{-x} - \frac{e^{-2x}}{2} + \frac{e^{-3x}}{3} - \frac{e^{-4x}}{4} + \cdots. $$
This assumes $x \geq 0$. If $x \leq 0$ you should use
$$ \log(1+e^x) = e^x - \frac{e^{2x}}{2} + \frac{e^{3x}}{3} - \frac{e^{4x}}{4} + \cdots. $$
Both series converge quickly only if $|x|$ is large.
A: Manipulations of this and similar quantities like $ \frac{e^x}{1+e^x}$, or $e^{-x} (1 + e^x)$, and their logarithms, are common in logistic regression in statistics. 
There is no closed form simplification but you can expand $\log(1 \pm t) = \pm t + \frac{t^2}{2} \pm \frac{t^2}{3} + \cdots$ in a power series for specially chosen $t$ (smaller than 1 in absolute value) to illustrate how $f(x) = \log (1 + e^x)$ is approximately $x$. Possibilities include


*

*to write $f(x) = x + \log (1 + e^{-x})$ and use $t = e^{-x}$ to get a series valid for positive $x$, 

*or $t = \frac{1}{1+e^x}$, for which $\log (1 + e^x) = x - \log(1-t) = x + t + \frac{t^2}{2} + \frac{t^3}{3} + \cdots$,  is valid for all $x$.
