So, here is the function: $$\frac{1}{1+e^{-x}}$$, for $-\infty<x<\infty$. I need to make sure that the given function is PDF. So, I must check whether
$$\int^{\infty}_{-\infty}\frac{1}{1+e^{-x}}dx\stackrel{?}{=}1$$
$$\int \frac{1}{1+e^{-x}}dx=\left[ u=1+e^{-x} \rightarrow du=-e^{-x} \rightarrow dx=-e^x \right]= \int \frac{1}{(1-u)u}du=\int\frac{1}{u-u^2}du=ln(u)+\frac{1}{u}=ln(1+e^{-x})+\left.\frac{1}{1+e^{-x} }\right|^{\infty}_{-\infty}=lim_{x \to \infty}\left(ln(1+e^{-x})+\frac{1}{1+e^{-x}} \right)-lim_{x\to -\infty}\left( ln(1+e^{-x})+\frac{1}{1+e^{-x}} \right)=1-\infty$$
The result seems to be off. can someone tell me where is my mistake?
Best Answer
By showing that $\lim_{x\rightarrow -\infty}\frac{1}{1+e^{-x}}=0$, $\lim_{x\rightarrow \infty}\frac{1}{1+e^{-x}}=1$ and $\frac{\partial}{dx}\frac{1}{1+e^{-x}}\geq0$ you can show that the function is a cumulative distribution function (not a pdf)