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.
 A: 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.
A: 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}
A: If $k$ is an integer multiple of $s^2$, then it appears you can use the result $I(0)=\frac12$ to obtain $I(k)$ as a sum of a finite number of terms.
