If $X\sim \exp(1)$, for $x \in \{0,+\infty \}$, and $Y = \frac{1}{1 + e^{aX} }$, where $a>0$ is a constant. (So the limit of Y becomes $y \in \{0, 1/2\}$) - Is it correct?
CDF $F_Y(y)$ of Y is :
$$F_Y(y) = \mathbb{P}(Y\leq y)= \Big(\frac{1}{1 + e^{aX}} \leq y \Big)$$
after mathematical manipulations, I get
$$F_Y(y) = \frac{1}{(y-1)^{1/a}}$$
So the PDF is:
$$f_Y(y) = \frac{1}{a}(y-1)^{-\frac{a-1}{a}}$$
The expected value of Y is now
$$\mathbb{E}[Y] = \int_0^{1/2} y\times\frac{1}{a}(y-1)^{-\frac{a-1}{a}} dy $$
$$\mathbb{E} = \dfrac{a\left(2a\mathrm{e}^\frac{\ln\left(-\frac{1}{2}\right)}{a}+\left(1-2a\right)\mathrm{e}^\frac{\ln\left(-1\right)}{a}\right)\mathrm{e}^{-\frac{\ln\left(-\frac{1}{2}\right)}{a}-\frac{\ln\left(-1\right)}{a}}}{2\left(a-1\right)}$$
It is more direct to compute $$\begin{align} \operatorname{E}[Y] &= \operatorname{E}[(1+e^{aX})^{-1}] \\ &= \int_{x=0}^\infty \frac{1}{1+e^{ax}} f_X(x) \, dx \\ &= \int_{x=0}^\infty \frac{e^{-x}}{1+e^{ax}} \, dx \\ &= \int_{x=0}^\infty e^{-x} \sum_{k=1}^\infty (-1)^{k-1} e^{-k a x} \\ &= \sum_{k=1}^\infty (-1)^{k-1} \int_{x=0}^\infty e^{-(ka+1) x} \, dx \\ &= \sum_{k=1}^\infty \frac{(-1)^{k-1}}{ka+1}, \end{align}$$ for $a > 0$. If $a < 0$, then the series expansion is modified accordingly.