Prove $E[\frac{1}{1+e^{t X}}] = 1/2$ for $X\sim N(0,1)$. I'm trying to evaluate an expectation of the form
$$
E\left[\frac{\exp(t X_1)}{\sum_{i=1}^n \exp(t X_i)}\right],
$$
where each $X_i$ is independent standard normally distributed.
In the specific case $n=2$ this can be rewritten as
$$E\left[\frac{1}{1+e^{\sqrt{2} t X}}\right],$$
which numerically appears to be equal to $1/2$ for all real $t$.
I wonder if that might be possible to prove?, and if it could shed some light on the original integral?

If I series expand I get the sum
$$E\left[\frac{1}{1+e^{t X}}\right]
=\sum_{k\ge 0}(-1)^k e^{(kt)^2/2},
$$
which Mathematica evaluates as $\frac{1}{2} \left(\vartheta _4\left(0,e^{\frac{t^2}{2}}\right)+1\right)$.
I don't really know what this EllipticTheta[4, 0, E^(t^2/2)] function is, but if it is 0 it at least fits with my conjecture.

Let $M = E\left[\frac{\exp(t X_1)}{\sum_{i=1}^n \exp(t X_i)}\right]$.  Since $X_j$'s are IID, so $M = E\left[\frac{\exp(t X_j)}{\sum_{i=1}^n \exp(t X_i)}\right]$ for all $j = 1,\dots,n$.  Take summation over $j = 1,\dots,n$.  We get $nM = E[1] = 1$, so $M = 1/n$.
