Originally Posted by

**evidor** How would you show that the expected value of the logistic distribution is equal to it's mean? ie E(X)=mu

$\displaystyle E(X) = \int_{-\infty}^{\infty} {\frac{xe^{-(x-\mu)/s}} {s\left(1+e^{-(x-\mu)/s}\right)^2}} \!$

slightly simplified:

$\displaystyle E(X) = e^{\mu/s}/s \int_{-\infty}^{\infty} {\frac{xe^{-x/s}} {\left(1+e^{-(x-\mu)/s}\right)^2}} \!$

Various substitutions I've tried don't seem to work, $\displaystyle u={1+e^{-(x-\mu)/s}}$, etc haven't worked to well for me.