pdf: $ f(x) = \frac{e^{-x}}{(1+e^{-x})^2} $
Find the mean of the distribution using: $ \int_{-\infty}^{\infty} xf(x) \ \mathrm{d}x $
Is it possible?
I have run the integral in a calculator only to find out that it diverges.
I got: $ \int x\frac{e^{-x}}{(1+e^{-x})^2} \ \mathrm{d}x = \frac{x}{1+e^{-x}}-\ln \left|e^{-x}+1\right|+C $
which should basically diverge. Am I missing something here?
First of all, notice that $|e^{-x}+1| = e^{-x}+1.$ Moreover, observe that for any $a > 0$:
$$\int_{-a}^a x\frac{e^{-x}}{(1+e^{-x})^2} \ \mathrm{d}x = -\left[\log(e^{-x} + 1) + \frac{x}{e^{x} + 1}\right]_{x=-a}^{x=a} = \\ -\left[\log(e^{-a} + 1) + \frac{a}{e^{a} + 1} - \log(e^{a} + 1) - \frac{-a}{e^{-a} + 1}\right] = \\ -\left[\log\left(\frac{e^{-a} + 1}{e^{a} + 1}\right) + \frac{a}{e^{a} + 1} + \frac{a}{e^{-a} + 1}\right] = \\ -\left[\log\left(\frac{e^{-a}(e^{a} + 1)}{e^{a} + 1}\right) + \frac{a}{e^{a} + 1} + \frac{ae^{a}}{e^{a}(e^{-a} + 1)}\right] = \\ -\left[\log e^{-a} + \frac{a}{e^{a} + 1} + \frac{ae^{a}}{e^{a} + 1}\right] = \\ -\left[-a + \frac{a(e^a + 1)}{e^{a} + 1} \right] = -[-a+a] = 0.\\ $$