Integral of product of gaussian CDF and PDF Looking for an analytic solution to the integral below:
$$
\int_{-\infty}^\infty \Phi\left(\frac{x - a}{\tau}\right) \phi\left(\frac{x - b}{\sigma}\right)dx
$$
where $\Phi(\cdot)$ and $\phi(\cdot)$ are, respectively, the standard normal CDF and PDF.
 A: These notes are just intended as an extended comment on Iosif Pinelis' answer.  In my initial version of this post, I made a mistake, but Iosif pointed out the error in the comments below; it should now be correct (and in agreement with Iosif's accepted answer).
Let $I$ be the original poster's integral.  Let $C\sim \mathcal{N}(a,\,\tau^{2}), D\sim \mathcal{N}(b,\,\sigma^{2})$ with $C,D$ independent.  Then
$$ \Pr[C<D] = \int_{-\infty}^\infty \Phi\left(\frac{x - a}{\tau}\right) \left[\frac{1}{\sigma} \phi\left(\frac{x - b}{\sigma}\right)\right]dx=I/\sigma$$
However, since $C$ and $D$ are independent, we have that $C-D \sim \mathcal{N}(a-b,\,\sigma^2 + \tau^{2})$.  Then
$$ \Pr[C<D] = \Pr[C-D<0] = \Phi\Big(\frac{b-a}{\sqrt{\tau^2+\sigma^2}}\Big)$$
Combining the above,
$$ I = \sigma \Phi\Big(\frac{b-a}{\sqrt{\tau^2+\sigma^2}}\Big)$$
A: $\newcommand\si\sigma$Let $I$ denote the integral in question. Then
$$I=\si\Phi\Big(\frac{b-a}{\sqrt{\tau^2+\si^2}}\Big). $$
Indeed, using the substitution $x=b+\si u$ and letting
$$A:=\frac\si\tau,\quad B:=\frac{b-a}\tau,$$
we have
$$I=\si\int_{-\infty}^\infty\Phi(Au+B)\phi(u)\,du
=\si P(V\le AU+B),$$
where $U,V$ are independent standard normal random variables. So, for $W:=V-AU$ we have $W\sim N(0,1+A^2)$ and hence
\begin{align*}
I=\si P(V\le AU+B)&=\si P(W\le B) \\
&=\si\Phi\Big(\frac B{\sqrt{1+A^2}}\Big)
=\si\Phi\Big(\frac{b-a}{\sqrt{\tau^2+\si^2}}\Big),\tag{1}\label{1}
\end{align*}
as claimed.

There has been a (strange to me) assertion in the discussion of my answer  that somehow the value of $I$ should be $\le1$ -- and actually a number of users seem to have been receptive to this assertion. Neither my answer itself nor the arguments in my comments about this assertion seem to have had any effect; on the other hand, no mistake in my arguments has been indicated.
So, I am now giving another argument, involving Mathematica. Strangely, Mathematica cannot evaluate the integral in general, even in the simple case when $a=b$ but $\tau$ is an arbitrary positive real number. However, Mathematica can evaluate the integral when $a=b=0$ and $\tau=1$:

Of course, this is fully consistent with formula \eqref{1}. In particular, it follows that $I>1$ if $\si>2$ (given that $a=b=0$ and $\tau=1$).
A: \begin{align}
& \sigma \int_{-\infty}^{+\infty} \Phi\left(\frac{x - a}{\tau}\right) \overbrace{ \varphi\left(\frac{x - b}{\sigma}\right) \frac{dx} \sigma }^{ \begin{smallmatrix} & \text{This is the} \\ & \text{$\operatorname N(b,\sigma^2)$ distribution.} \end{smallmatrix} } \\[10pt]
= {} & \sigma\operatorname E\left( \Phi\left( \frac{X-a}\tau \right) \right) \text{ where } X\sim\operatorname N(b,\sigma^2) \\[10pt]
= {} & \sigma\operatorname E\left( \Pr\left(Z\le \frac{X-a} \tau \right) \,\Bigg\vert\, X \right) \\
& \text{where } Z\sim\operatorname N(0,1) \text{ and } X,Z \text{ are independent} \\[10pt]
= {} & \sigma\Pr\left( Z\le \frac{X-a} \tau \right)  \\
& \text{(law of total probability)} \\[10pt]
= {} & \sigma\Pr\left( Z - \frac{X-a} \tau \le0 \right) \\[10pt]
= {} & \sigma\Pr(W\le 0) \text{ where } W\sim\operatorname N\left( \frac{a-b}\tau, 1 + \frac{\sigma^2}{\tau^2} \right) \\[10pt]
= {} & \sigma\Pr\left( \frac{W- \frac{a-b}\tau}{\sqrt{1+\frac{\sigma^2}{\tau^2}}} \le \frac{0- \frac{a-b}\tau}{\sqrt{1+\frac{\sigma^2}{\tau^2}}} \right) \\[10pt]
= {} & \sigma \Pr\left( Z \le \frac{0- \frac{a-b}\tau}{\sqrt{1+\frac{\sigma^2}{\tau^2}}} \right) = \sigma \Pr\left( Z\le \frac{b-a}{\sqrt{\tau^2+\sigma^2}} \right) \\[10pt]
=  {} & \sigma \Phi\left( \frac{b-a}{\sqrt{\tau^2+\sigma^2}} \right).
\end{align}
