By the algebra below I keep getting $F=\frac{1}{F}$, where $F$ is the log function. I appreciate if someone can point out the error.
$F(t)=\frac{1}{1+e^{-t}}$ $f(t)=\frac{dF}{dt}=\frac{e^{-t}}{(1+e^{-t})^2}$ $\frac{f}{F}=\frac{1}{1+e^t}$ $\frac{f}{F}=\frac{d\ln{F}}{dt} \rightarrow \int d\ln{F}=\int \frac{1}{1+e^t} dt $ $\ln{F}=\ln(1+e^{-t})+C$ $F=1+e^{-t}$ Looking back at the first equation, this implies $F=\frac{1}{F}$