Evaluating $\int_0^\infty\Phi\left(\frac{\ln x-\mu}{\sigma}\right)\left(1-\Phi\left(\frac{\ln x-\mu}{\sigma}\right)\right)\mathrm{d}x$ I am trying to evaluate the following integral where $\Phi$ denotes the CDF of $\mathcal N(0,1)$:
$$I=\int_0^\infty\Phi\left(\frac{\ln x-\mu}{\sigma}\right)\left(1-\Phi\left(\frac{\ln x-\mu}{\sigma}\right)\right)\,\mathrm{d}x$$
$$=\sigma e^{\mu}\int_{-\infty}^\infty \Phi(t)(1-\Phi(t))e^{\sigma t}\,\mathrm{d}t$$
The integral arises while computing the Gini coefficient of concentration for the Lognormal distribution. Apparently, the answer should turn out to be $\left(2\Phi\left(\frac{\sigma}{\sqrt 2}\right)-1\right)e^{\mu+\sigma^2/2}$.
I tried integrating by parts taking $e^{\sigma t}$ as the second function but if I do that the integral seems to diverge. I must be missing something obvious. I can certainly write the integral as $\displaystyle I=\sigma e^{\mu}\int_{-\infty}^\infty \Phi(t)\Phi(-t)e^{\sigma t}\,\mathrm{d}t$ also, but I don't think that helps. Any suggestion would be great.
 A: Consider $(X,Y)$ i.i.d. standard normal, then, for every $u$, 
$$
\Phi(u)(1-\Phi(u))=P(X\leqslant u)P(u<Y)=P(X\leqslant u<Y)
$$ 
hence 
$$
I=\int_0^\infty P\left(X\leqslant\frac{\log x-\mu}\sigma<Y\right)dx=\int_0^\infty P\left(e^{\sigma X+\mu}\leqslant x<e^{\sigma Y+\mu}\right)dx
$$ 
that is, 
$$
I=E\left(\int_0^\infty \mathbf 1_{e^{\sigma X+\mu}\leqslant x<e^{\sigma Y+\mu}}dx\right)=E(e^{\sigma Y+\mu}-e^{\sigma X+\mu};X\leqslant Y)
$$ 
Now, 
$$
E(e^{\sigma Y};Y\geqslant x)=\int_x^\infty e^{\sigma y}\varphi(y)dy=e^{\sigma^2/2}\int_x^\infty \varphi(y-\sigma)dy=e^{\sigma^2/2}(1-\Phi(x-\sigma))
$$ 
and
$$
E(e^{\sigma X};X\leqslant y)=\int_{-\infty}^y e^{\sigma x}\varphi(x)dx=e^{\sigma^2/2}\int_{-\infty}^y \varphi(x-\sigma)dx=e^{\sigma^2/2}\Phi(y-\sigma)
$$
hence 
$$
I=e^{\mu}e^{\sigma^2/2}E(1-\Phi(X-\sigma))-e^{\mu}e^{\sigma^2/2}E(\Phi(Y-\sigma))=e^{\mu}e^{\sigma^2/2}(1-2E(\Phi(X-\sigma))$$ 
Note that 
$$
E(\Phi(X-\sigma))=P(Y\leqslant X-\sigma)=P(Y-X\leqslant-\sigma)=\Phi(-\sigma/\sqrt2)=1-\Phi(\sigma/\sqrt2)
$$ 
because $Y-X$ is centered normal with variance $2$. Finally,

$$I=e^{\mu+\sigma^2/2}(2\Phi(\sigma/\sqrt2)-1)$$

