A hard integral

The change of variable $v = \log u$ shows that you're trying to integrate the logistic-normal integral.

$$\int_{-\infty}^{\infty} \frac{e^{-\frac{1}{2}\left(\frac{v-k}{s}\right)^2}}{\sqrt{2\pi} s} \frac{1}{1+e^{-v}}~\mathrm{d}v$$

I doubt there is a closed form solution, and none seems known.

See http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.372.3781&rep=rep1&type=pdf for the approximation

$$\left|I(s,k)- \frac{1}{1+e^{-\frac{k}{\sqrt{1+\frac{\pi s^2}{8}}}}}\right| < 0.02$$

and http://www.sciencedirect.com/science/article/pii/S0377042712002518 for a deeper discussion.

Here is a start: $I(0) = \frac{1}{2}$

Proof:

$$I(0) = \int\limits_0^\infty \frac{\exp\left[-\frac{(\log u)^2}{2s^2}\right]}{\sqrt{2\pi} s (1+u)} \rm{d}u$$

Put $\log u = x$

\begin{align} I(0) &= \int\limits_{-\infty}^\infty \frac{\exp\left[-\frac{x^2}{2s^2}\right]}{\sqrt{2\pi} s} \frac{e^x}{1+e^x} \rm{d}x \\ &= \int\limits_{-\infty}^\infty \frac{\exp\left[-\frac{x^2}{2s^2}\right]}{\sqrt{2\pi} s}\rm{d}x - \int\limits_{-\infty}^\infty \frac{\exp\left[-\frac{x^2}{2s^2}\right]}{\sqrt{2\pi} s} \frac{1}{1+e^x} \rm{d}x \end{align}

The first integral is $1$. Call the second integral $K$. $$K=\int\limits_{-\infty}^\infty \frac{\exp\left[-\frac{x^2}{2s^2}\right]}{\sqrt{2\pi} s} \frac{1}{1+e^x} \rm{d}x$$

Flipping the range around $0$, $$K=\int\limits_{-\infty}^\infty \frac{\exp\left[-\frac{x^2}{2s^2}\right]}{\sqrt{2\pi} s} \frac{1}{1+e^{-x}} \rm{d}x$$

Now take the average of the two expressions, \begin{align} K &=\frac{1}{2}\int\limits_{-\infty}^\infty \frac{\exp\left[-\frac{x^2}{2s^2}\right]}{\sqrt{2\pi} s} \left[\frac{1}{1+e^x}+\frac{1}{1+e^{-x}}\right] \rm{d}x\\ &=\frac{1}{2}\int\limits_{-\infty}^\infty \frac{\exp\left[-\frac{x^2}{2s^2}\right]}{\sqrt{2\pi} s}\rm{d}x\\ &=\frac{1}{2}\\ I(0) &= 1 - K = \frac{1}{2} \end{align}