Evaluating an integral with standard normal distribution as integrand Consider
\begin{equation}
\int_{0}^{\infty}\frac{1 - e^{-su}}{u}\Phi(-\sqrt{2u})du,
\end{equation}
where $s > 0$ is a real number and $\Phi$ is the distribution function of the standard normal distribution, i.e.
\begin{equation}
\Phi(-\sqrt{2u}) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{-\sqrt{2u}}e^{-\frac{1}{2}t^{2}}dt.
\end{equation}
The solution is given by 
\begin{equation}
\log(\frac{1 + \sqrt{1+s}}{2})
\end{equation}
But I don't know how get the solution. Can anyone help me
 A: A common trick to try is to take the derivative with respect to the parameter $s$:
$$\frac{d}{ds}\int_{0}^{\infty}\frac{1 - e^{-su}}{u}\Phi(-\sqrt{2u})du=\int_{0}^{\infty}e^{-su}\Phi(-\sqrt{2u})du$$
Then you can perform a partial integration
$$\begin{eqnarray}-\frac{1}{s}\int_{u=0}^{u=\infty}\Phi(-\sqrt{2u})de^{-su} & = & \left[-\frac{1}{s}e^{-su}\Phi(-\sqrt{2u})\right]_{u=0}^{u=\infty}+\frac{1}{s}\int_{u=0}^{u=\infty}e^{-su}d\Phi(-\sqrt{2u}) \\ & = & \frac{1}{2s}+\frac{1}{\sqrt{2}s}\int_{u=0}^{u=\infty}e^{-su}\phi(-\sqrt{2u})(-u^{-1/2})du\end{eqnarray}$$
Substituting the formula for the density of the normal distribution $\phi$ we get
$$\begin{eqnarray}\frac{1}{2s}+\frac{1}{\sqrt{2}s}\int_{0}^{\infty}e^{-su}\phi(-\sqrt{2u})(-u^{-1/2})du & = & \frac{1}{2s}-\frac{1}{2\sqrt{\pi}s}\int_{0}^{\infty}e^{-su}e^{-u}u^{-1/2}du \\ & = & \frac{1}{2s}-\frac{1}{2\sqrt{\pi}s\sqrt{s+1}}\Gamma(1/2) \\ & = & \frac{1}{2s}-\frac{1}{2s\sqrt{s+1}}\end{eqnarray}$$
So the derivative of our integral is equal to that last part. If we integrate it, by for instance using a substitution $x=\sqrt{s+1}$, we get
$$\log(1+\sqrt{1+s})+c$$
The integration constant can be fixed by noting that our initial integral is $0$ when $s=0$, therefore $c=-\log(2)$.
