Does $ \int_{-\infty}^{\infty}\Phi(x)\phi\left(\frac{x-a}{b}\right)\,dx $ have a closed form solution? Is there a solution to the following Integral that can get rid of the integral, but can leave the answer in terms of $\Phi()$ (the standard normal CDF) and $\phi $ (the standard normal PDF).?
$$ \int_{-\infty}^{\infty}\Phi(x)\phi\left(\frac{x-a}{b}\right)\,dx  $$
where $a, b$ are constants with $b>0$,
I've tried several variants of integration by parts, all of which get me nowhere. I'm thinking the likely answer is no.
 A: The integral in question can be given an probabilistic interpretation:
$$ \begin{split}
   \int_\mathbb{R} \Phi(x) \phi\left(\frac{x-a}{b}\right) \mathrm{d}x &= b \int_\mathbb{R} \Pr(Z \leqslant x) \phi\left(\frac{x-a}{b}\right) \frac{\mathrm{d}x}{b} \\ &= b \int_\mathbb{R} \Pr(Z \leqslant a+b y) \phi\left(y\right) \mathrm{d}y \\ &=
   b \Pr\left(Z \leqslant a + b Y \right) = b \Pr\left(Z - b Y \leqslant a \right)
\end{split} $$
Since $Z$ and $Y$ are both independent standard normal random variables, $Z-b Y$ is also a normal random variables, with mean and variance:
$$
    \mathbb{E}(Z-b Y) = 0, \quad \mathbb{Var}(Z-b Y) = \mathbb{Var}(Z) + b^2 \mathbb{Var}(Y) = 1+b^2
$$
Thus $Z- b Y \stackrel{d}{=} X$, where $X \sim \mathcal{N}\left(0, \sqrt{1+b^2} \right)$ we have:
$$
  \int_\mathbb{R} \Phi(x) \phi\left(\frac{x-a}{b}\right) \mathrm{d}x = b \Pr\left(Z - b Y \leqslant a \right) = b \Pr\left(X \leqslant a \right) = b \, \Phi\left(\frac{a}{\sqrt{1+b^2}}\right) \tag{1}
$$
Numerical check of $(1)$:
In[25]:= With[{a = -3., 
  b = 2.3}, {NIntegrate[
   CDF[NormalDistribution[], x] PDF[
     NormalDistribution[], (x - a)/b], {x, -Infinity, Infinity}], 
  b CDF[NormalDistribution[], a/Sqrt[1 + b^2]]}]

Out[25]= {0.266371, 0.266371}

