Limit of an integral involving the normal CDF Let $\Phi $ denote the standard normal CDF, and $\phi$ the standard normal PDF. Fix $\alpha > 0$.
Let 
$$ Z\left( r\right) =r\int_{0}^{\infty } e^{-(r+\alpha )t} \mathbb{E} \left[ \Phi
\left( \frac{e^{-\alpha t}(\theta -\omega )}{\sqrt{\frac{1%
}{2}(1-e^{-2\alpha t})}}\right)  \frac{1-\Phi (\theta )}{\phi(\theta)}  \; \right] \; dt $$
The expectation is over the random variables $\theta $ and $\omega$ and these two variables are drawn from the standard normal distribution, independently. 
Question: What is the following limit? $$ \lim_{r\rightarrow 0^{+}}Z( r ) $$ 
Low-quality numerical evidence seems to suggest zero is the answer, but this is only a hunch, and I have not been able to prove it.
 A: Let $U$, $X$ and $Y$ denote independent random variables with $U$ uniform on $(0,1)$ and $X$ and $Y$ standard normal. For every positive $s$, introduce
$$
K(s)=\mathbb{E}\left[U^{s}\Phi\left(W(X-Y)\right)\frac{1-\Phi(X)}{\phi(X)}\right],\qquad
W=\frac{U}{\sqrt{\frac12(1-U^2)}}.
$$
Some simple computations show that for every positive $r$ and $\alpha$,
$$
Z(r)=K(r/\alpha)r/\alpha.
$$ 
Hence, when $r\to0^+$, $Z(r)$ converges to the limit of $sK(s)$ when $s\to0^+$. We now study this limit.

First step: rewriting $K(s)$ Let $Z$ denote a standard normal random variable independent from $Y$, $w$ a positive real number and $x$ a real number. Then
$$
\mathbb{E}\left[\Phi\left(w(x-Y)\right)\right]=
\mathbb{P}\left[Z\le w(x-Y)\right]=
\mathbb{P}\left[Y+\frac{1}{w}Z\le x\right].
$$
Since $\displaystyle Y+\frac{1}{w}Z$ is a centered normal random variable with variance $\displaystyle 1+\frac1{w^2}$,
$$
\mathbb{E}\left[\Phi\left(w(x-Y)\right)\right]=\Phi\left(xv\right),\qquad \frac1{v^2}=1+\frac1{w^2}.
$$
Integrating this relation over $W$ and $X$, one gets
$$
K(s)=\mathbb{E}\left[U^{s}\Phi\left(XV\right)\frac{1-\Phi(X)}{\phi(X)}\right],
\qquad V=\sqrt{\frac{2U^2}{1+U^2}}.
$$
Let us write $K(s)$ as $K(s)=K_+(s)+K_-(s)$ where $K_+(s)$ corresponds to the expectation on $X\ge0$ and $K_-(s)$ to the expectation on $X\le0$.

Second step: behaviour of $K_+(s)$ Note that, if $X\ge0$, then $U^s\Phi(XV)\le1$, hence
$$
K_+(s)\le\mathbb{E}\left[\frac{1-\Phi(X)}{\phi(X)};X\ge0\right]=\int_0^{+\infty}\mathrm{d}x\int_{x}^{+\infty}\phi(y)\mathrm{d}y=\int_0^{+\infty}y\phi(y)\mathrm{d}y=\frac1{\sqrt{2\pi}}.
$$

Third step: behaviour of $K_-(s)$ Note that, if $X\le0$, then $1/2\le 1-\Phi(X)\le1$, hence $L(s)/2\le K_-(s)\le L(s)$, with
$$
L(s)=\mathbb{E}\left[U^{s}\Phi\left(XV\right)\frac1{\phi(X)};X\le0\right]
=\mathbb{E}\left[U^{s}(1-\Phi\left(XV\right))\frac1{\phi(X)};X\ge0\right].
$$
For every positive $v$,
$$
\mathbb{E}\left[(1-\Phi\left(Xv\right))\frac1{\phi(X)};X\ge0\right]
=\int_0^{+\infty}\mathrm{d}x\int_{xv}^{+\infty}\phi(y)\mathrm{d}y
=\int_0^{+\infty}\frac{y}{v}\phi(y)\mathrm{d}y=\frac1{v\sqrt{2\pi}}.
$$
Hence,
$$
\sqrt{2\pi}L(s)=\mathbb{E}\left[U^{s}\frac1V\right].
$$
Using the upper bound $V\le\sqrt2U$, one gets
$$
2\sqrt{\pi}L(s)\ge\mathbb{E}\left[U^{s-1}\right]=\frac1s.
$$

Conclusion All this proves that
$$
\liminf Z(r)=\liminf sK_-(s)\ge\frac12\liminf sL(s)\ge\frac1{4\sqrt\pi},
$$ 
hence $Z(r)$ does not converge to zero. Finally, we mention that it is not much more difficult to show that in fact
$$
\lim Z(r)=\lim sK_-(s)=\lim sL(s)=\frac1{2\sqrt\pi}.
$$
