How to compute the variance of a truncated logistic distribution / this exponential integral? A random variable $X$ follows a truncated logistic distribution with truncation in 0, location parameter $m$, scale parameter $s$, and has density
$$
f(x) = \frac{1+e^{m/s}}{e^{m/s}} \times \frac{e^{-\frac{x-m}{s}}}{s\left(1+e^{-\frac{x-m}{s}}\right)^2}
$$ 
for $x \geq 0$, and 0 otherwise.
How can I compute the variance of this distribution?
If $m = 0, s = 1$, then following the answer here allows to compute the second moment.
However, for the general case, I have to compute 
$$
\int_0^\infty\frac{x^2e^{-\frac{x-m}{s}}}{s\left(1+e^{-\frac{x-m}{s}}\right)^2}dx,
$$
Substitution $z = \frac{x-m}{s}$ leads to a similar looking integral,
$$
\int_{-\frac{m}{s}}^\infty\frac{(zs+m)^2 e^{-z}}{\left(1+e^{-z}\right)^2} dz,
$$
where, however, I don't know how to proceed due to the shift in integration limit.
How can I compute the mean value?
 A: Don't know if this helps, after the long time -- i had a similar question, and this is what I got  from Mathematica.  Would welcome any tips on simplifying it
d = TruncatedDistribution[{a, b}, LogisticDistribution[m, s]]
FullSimplify[Variance[d], a < b]
$-\frac{e^{-\frac{2 m}{s}} \left(e^{a/s}+e^{m/s}\right) \left(e^{b/s}+e^{m/s}\right) \left(2 s^2 e^{m/s} \left(e^{a/s}-e^{b/s}\right) \left(\text{Li}_2\left(-e^{\frac{a-m}{s}}\right)-\text{Li}_2\left(-e^{\frac{b-m}{s}}\right)\right)+2 a s e^{m/s} \left(e^{a/s}-e^{b/s}\right) \log \left(e^{\frac{a-m}{s}}+1\right)-2 b s e^{m/s} \left(e^{a/s}-e^{b/s}\right) \log \left(e^{\frac{b-m}{s}}+1\right)+s \left(-2 e^{m/s} \left(a e^{a/s}-b e^{b/s}\right)+s \left(e^{a/s}+e^{m/s}\right) \left(e^{b/s}+e^{m/s}\right) \left(\log \left(e^{a/s}+e^{m/s}\right)-\log \left(e^{b/s}+e^{m/s}\right)\right)-2 (a-b) e^{\frac{a+b}{s}}\right) \left(\log \left(e^{a/s}+e^{m/s}\right)-\log \left(e^{b/s}+e^{m/s}\right)\right)+(a-b)^2 e^{\frac{a+b}{s}}\right)}{\left(e^{a/s}-e^{b/s}\right)^2}$
Here, $Li_2$ is the polylog function
