A hard integral Looking for a solution for an integral:
$$I(k)=\int_0^{\infty } \frac{e^{-\frac{(\log (u)-k)^2}{2 s^2}}}{\sqrt{2 \pi } s \left(1+u\right)} \, du .$$
So far I tried substitutions and by parts to no avail.

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}
