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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 $$F=\frac{1}{F}$$

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I don't understand what's going on towards the end. You are integrating $1/(1+e^t)$, which is $e^{-t}/(1+e^{-t})$. You assert that the integral is $\ln(1+e^{-t})+C$. But that's not true, if you use substitution $u=1+e^{-t}$, then $du=-e^{-t}\,dt$. –  André Nicolas Nov 4 '12 at 22:11
    
@Ben: My correction was incorrect. You should roll it back. –  Joseph Quinsey Nov 4 '12 at 23:29

3 Answers 3

up vote 2 down vote accepted

$$\int \frac{1}{1+e^t} dt=-\ln(1+e^{-t})+C,\quad \text{not}\ +\ln(1+e^{-t})+C$$

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I think you're missing a $t$. –  nbubis Nov 4 '12 at 22:14

Towards the end, you are integrating $1/(1+e^t)$, which is $e^{-t}/(1+e^{-t})$. You assert that the integral is $\ln(1+e^{-t})+C$. But that's not true. If you use substitution $u=1+e^{-t}$, then $du=-e^{-t}\,dt$, so the integral should be $-\ln(1+e^{-t})+C$.

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You goofed in your integration. Use the substitution $\mu=1+e^{-t}$, so that $$\int\frac1{1+e^t}\,dt=-\int\frac{-e^{-t}}{1+e^{-t}}\,dt=-\int\frac1{\mu}\,d\mu=-\ln\mu+C=\ln\frac1\mu+C=\ln\frac1{1+e^{-t}}+C.$$

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