\[\begin{align*} E(X) &= \int_{-\infty}^\infty x\frac{e^{-(x-\mu)/s}}{s(1+e^{-(x-\mu)/s})^2}dx\\ &= \frac{1}{s}\int_{-\infty}^\infty \frac{xe^{-(x-\mu)/s}}{(1+e^{-(x-\mu)/s})^2}dx\\ \text{Let: } u &= \frac{1}{1+\exp\left(-\frac{x-\mu}{s} \right)} \\ \therefore \frac{du}{dx} &= \frac{1}{(1+\exp\left(-\frac{x-\mu}{s} \right))^2}\left(-\frac{1}{s}\exp\left(-\frac{x-\mu}{s}\right)\right)\\ \Rightarrow dx &= \frac{s\left(1+\exp\left(-\frac{x-\mu}{s} \right)\right)^2}{\exp\left(-\frac{x-\mu}{s}\right)} du \\ \text{and:} \frac{1}{u} - 1 &= \exp\left(-\frac{x-\mu}{s} \right) \\ \ln\left(\frac{1}{u} - 1 \right) &= -\frac{x-\mu}{s} \\ \Rightarrow x &= -s\ln\left(\frac{1-u}{u} \right) + \mu \\ \text{and:} \lim_{x\to -\infty} \left(1+\exp\left(-\frac{x-\mu}{s}\right)^{-1} \right) &=0 \\ \lim_{x\to \infty} \left(1+\exp\left(-\frac{x-\mu}{s}\right)^{-1} \right) &=1 \\ E(X) &= \int_{\to 0}^{\to 1} \left(-s\ln\left(\frac{1-u}{u} \right) + \mu \right) du \\ &= -s\int_{\to 0}^{\to 1} \ln\left(\frac{1-u}{u} \right) du + \mu \int_{\to 0}^{\to 1} du \\ &= -s\left[\int_{\to 0}^{\to 1} \ln(1-u) du + \int_{\to 0}^{\to 1} \ln(u) du \right] + \mu\\ &= -s[(1-u)\ln(1-u) - (1-u)]_0^1 + s[u\ln u - u]_0^1 + \mu \\ &= -s(0 - \ln(1) - 1) + s(\ln (1) - 1) + \mu \\ &= \mu \end{align*}\]
\[\begin{align*} E(X^2) &= \int_{-\infty}^\infty x^2\frac{e^{-(x-\mu)/s}}{s(1+e^{-(x-\mu)/s})^2}dx\\ &= \frac{1}{s}\int_{-\infty}^\infty \frac{x^2e^{-(x-\mu)/s}}{(1+e^{-(x-\mu)/s})^2}dx\\ \text{Let: } u &= \frac{1}{1+\exp\left(-\frac{x-\mu}{s} \right)} \\ \therefore \frac{du}{dx} &= \frac{1}{(1+\exp\left(-\frac{x-\mu}{s} \right))^2}\left(-\frac{1}{s}\exp\left(-\frac{x-\mu}{s}\right)\right)\\ \Rightarrow dx &= \frac{s\left(1+\exp\left(-\frac{x-\mu}{s} \right)\right)^2}{\exp\left(-\frac{x-\mu}{s}\right)} du \\ \text{and:} \frac{1}{u} - 1 &= \exp\left(-\frac{x-\mu}{s} \right) \\ \ln\left(\frac{1}{u} - 1 \right) &= -\frac{x-\mu}{s} \\ \Rightarrow x &= -s\ln\left(\frac{1-u}{u} \right) + \mu \\ \text{and:} \lim_{x\to -\infty} \left(1+\exp\left(-\frac{x-\mu}{s}\right)^{-1} \right) &=0 \\ \lim_{x\to \infty} \left(1+\exp\left(-\frac{x-\mu}{s}\right)^{-1} \right) &=1 \\ E(X^2) &= \int_{\to 0}^{\to 1} \left(-s\ln\left(\frac{1-u}{u} \right) + \mu \right)^2 du \\ &= \int_{\to 0}^{\to 1} \left(s^2\ln^2\left(\frac{1-u}{u} \right) - 2\mu s \ln\left(\frac{1-u}{u} \right) + \mu^2 \right) du \\ &= \int_{\to 0}^{\to 1} s^2\ln^2\left(\frac{1-u}{u} \right)du - 2\mu s \int_{\to 0}^{\to 1} \ln\left(\frac{1-u}{u} \right)du + \int_{\to 0}^{\to 1} \mu^2 du \\ &= s^2\left[\int_{\to 0}^{\to 1} \ln^2(1-u)du + \int_{\to 0}^{\to 1} 2\ln(1-u)\ln udu +\int_{\to 0}^{\to 1} \ln^2 udu\right] - 2\mu s \bigg[\int_{\to 0}^{\to 1} \ln(1-u)du + \\ & \qquad \int_{\to 0}^{\to 1} \ln udu\bigg] + \mu^2 \\ &= s^2\left[2 - 2\left(2-\frac{\pi^2}{6} \right) + 2 \right]- 2\mu s [(-1)-(-1)] + \mu^2 \\ &= s^2\left[4 - 4 - \frac{\pi^2}{3} \right]+ \mu^2 \\ &= \frac{s^2\pi^2}{3} + \mu^2 \\ \mathrm{Var}(X) &= \frac{s^2\pi^2}{3} + \mu^2 - \mu^2 \\ &= \frac{s^2\pi^2}{3} \end{align*}\]
Properties of \(\ln\):
\[\begin{align*} \int_0^1 \ln^n u &= \int_0^1 (\ln u)^n = (-1)^n n! \\ \therefore \int_0^1 \ln^2 u &= (-1)^22! \\ &= 2 \end{align*}\]
\[\begin{align*} \text{Beta}(a,b) &= \int_0^1 u^{a-1}(1-u)^{b-1}du \\ \text{and: } \frac{\partial}{\partial a} u^{a-1} &= u^{a-1}\ln u \\ \frac{\partial}{\partial b} (1-u)^{b-1} &= (1-u)^{b-1} \ln(1-u) \\ \therefore \frac{\partial^2}{\partial a \partial b} \text{Beta}(a,b) &= \int_0^1 u^{a-1}(1-u)^{b-1} \ln u \ln(1-u) du \\ \text{let: } a&= b=1: \\ \text{Beta}(1,1) &= \int_0^1 \ln u \ln(1-u)du \\ \text{since: } u^0(1-u)^0 &=1 \\ \ln \text{Beta}(a,b) &= \ln \left(\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} \right) \\ &= \ln \Gamma(a) + \ln \Gamma(b) - \ln \Gamma(a+b) \\ \frac{\partial}{\partial a} \ln \text{Beta}(a,b) &= \varphi(a) - \varphi(a+b), \varphi \text{ is the Digamma function} \\ \therefore \frac{\partial \text{Beta}}{\partial a} &= \text{Beta}(a,b)[\varphi(a) - \varphi(a+b)] \\ \frac{\partial \text{Beta}}{\partial b} &= \text{Beta}(a,b)[\varphi(b) - \varphi(a+b)] \\ \frac{\partial^2 \text{Beta}}{\partial a\partial b} &= \text{Beta}(a,b)[(\varphi(a) - \varphi(a+b))(\varphi(b) - \varphi(a+b)) - \varphi'(a+b)], \varphi' \text{ is the Trigamma function} \\ \text{at: } a&=b=1: \\ \text{Beta}(1,1) &=1 \\ \varphi(1) &= -\gamma, \varphi(2) = 1-\gamma \\ \varphi'(2) &= \frac{\pi^2}{6} - 1 \\ \therefore \text{Beta}(1,1) &= \int_0^1 \ln u \ln(1-u)du = 1[(-\gamma - (1-\gamma))(-\gamma - (1-\gamma))-\frac{\pi^2}{6} + 1] \\ &= 2-\frac{\pi^2}{6} \end{align*}\]
\[\begin{align*} M_X(t) &= \int_{-\infty}^\infty e^{tx} \frac{e^{-(x-\mu)/s}}{s(1+e^{-(x-\mu)/s})^2}dx \\ &= \frac{1}{s}\int_{-\infty}^\infty e^{tx} \frac{e^{-(x-\mu)/s}}{(1+e^{-(x-\mu)/s})^2}dx \\ \text{let: } u &= \frac{1}{\left(1+e^{-(x-\mu)/s} \right)} \\ \therefore \frac{du}{dx} &= \frac{(-\frac{1}{s})e^{-(x-\mu)/s}}{\left(1+e^{-(x-\mu)/s}\right)^2} \\ dx &= \frac{-s\left(1+e^{-(x-\mu)/s}\right)^2}{e^{-(x-\mu)/s}} du \\ \text{and: } \frac{1}{u} - 1 &= e^{-(x-\mu)/s} \\ e^{-\mu/s}\frac{1}{u} - 1 &= e^{-\mu/s} e^{-(x-\mu)/s} \\ e^{-\mu/s}\frac{1}{u} - 1 &= e^{-x/s} \\ \left(e^{-\mu/s}\frac{1}{u} - 1\right)^{-st} &= \left(e^{-x/s}\right)^{-st} \\ e^{-\mu t}\left(\frac{1}{u} - 1\right)^{-st} &= e^{xt} \\ \text{and:} \lim_{x\to -\infty} \left(1+\exp\left(-\frac{x-\mu}{s}\right)^{-1} \right) &=0 \\ \lim_{x\to \infty} \left(1+\exp\left(-\frac{x-\mu}{s}\right)^{-1} \right) &=1 \\ \therefore M_X(t) &= e^{\mu t} \int_{\to 0}^{\to 1} \left(\frac{1-u}{u}\right)^{-st}du \\ &= e^{\mu t} \int_{\to 0}^{\to 1} (1-u)^{-st} u^{st}du \\ &=e^{\mu t} \text{Beta}(1-st, 1+st) \quad \text{for } |st| < 1 \end{align*}\]