Integral of the product of Normal density and cdf I am struggling with an integral pretty similar to one already resolved in MO (link: Integration of the product of pdf & cdf of normal distribution ). I will reproduce the calculus bellow for the sake of clarity, but I want to stress the fact that my computatons are essentially a reproduction of the discussion of the previous thread.
In essence, I need to solve:
$$\int_{-\infty}^\infty\Phi\left(\frac{f-\mathbb{A}}{\mathbb{B}}\right)\phi(f)\,df,$$
where $\Phi$ is cdf of a standard normal, and $\phi$ its density. $\mathbb{B}$ is a negative constant.
As done in the aforementioned link, the idea here is to compute the derivative of the integral with respect to $\mathbb{A}$ (thanks to Dominated Convergence Theorem, integral and derivative can switch positions). With this, 
\begin{align*}
\partial_A\left[\int_{-\infty}^\infty\Phi\left(\frac{f-A}{B}\right)\phi(f)\,df\right]&=\int_{-\infty}^\infty\partial_A\left[\Phi\left(\frac{f-A}{B}\right)\phi(f)\right]\,df=\int_{-\infty}^\infty-\frac{1}{B}\phi\left(\frac{f-A}{B}\right)\phi(f)\,df
\end{align*}
We note now that 
$$\phi\left(\frac{f-A}{B}\right)\phi(f)=\frac{1}{2\pi}\exp\left(-\frac{1}{2}\left[\frac{(f-A)^2}{B^2}+f^2\right]\right)=\exp\left(-\frac{1}{2B^2}\left[f^2(1+B^2)+A^2-2Af\right]\right)$$
$$=\frac{1}{2\pi}\exp\left(-\frac{1}{2B^2}\left[\left(f\sqrt{1+B^2}-\frac{A}{\sqrt{1+B^2}}\right)^2+\frac{B^2}{1+B^2}A^2\right]\right)$$
Finally, then,
$$\partial_A\left[\int_{-\infty}^\infty\Phi\left(\frac{f-A}{B}\right)\phi(f)\,df\right]$$
$$\ \ \ \ \ =-\frac{1}{\sqrt{2\pi}B}\exp\left(-\frac{A^2}{2(1+B^2)}\right)\frac{1}{2\pi}\int_{-\infty}^\infty\exp\left(-\frac{1}{2B^2}\left[f\sqrt{1+B^2}-\frac{A}{\sqrt{1+B^2}}\right]^2\right)\,df$$
and with the change of variable 
\begin{align}
\left[y\longmapsto f\frac{\sqrt{1+B^2}}{B}-\frac{A}{B\sqrt{1+B^2}}\Longrightarrow df=\frac{B}{\sqrt{1+B^2}}\,dy\right]
\end{align}
we get 
\begin{align}
\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty\exp\left(-\frac{1}{2B^2}\left[f\sqrt{1+B^2}-\frac{A}{\sqrt{1+B^2}}\right]^2\right)\,df=\frac{B}{\sqrt{1+B^2}}\int_{-\infty}^{\infty}\phi(y)\,dy=\frac{B}{\sqrt{1+B^2}}
\end{align}
This means that 
\begin{align}
\partial_A\left[\int_{-\infty}^\infty\Phi\left(\frac{f-A}{B}\right)\phi(f)\,df\right]&=-\frac{1}{\sqrt{2\pi}B}\exp\left(-\frac{A^2}{2(1+B^2)}\right)\frac{B}{\sqrt{1+B^2}}=-\frac{1}{\sqrt{1+B^2}}\phi\left(\frac{A}{\sqrt{1+B^2}}\right)
\end{align}
At this point, given that (as $\mathbb{B}$ is negative)
$$\Phi\left(\frac{f-A}{\mathbb{B}}\right)\phi(f)=0$$
when $\mathbb{A}\rightarrow-\infty$, the integral we are looking for is equal to
\begin{align}
\int_{-\infty}^{\mathbb{A}}-\frac{1}{\sqrt{1+\mathbb{B}^2}}\phi\left(\frac{A}{\sqrt{1+\mathbb{B}^2}}\right)\,dA
\end{align}
Again with the obvious change of variables
$$\left[y\longmapsto\frac{A}{\sqrt{1+\mathbb{B}^2}}\Longrightarrow\sqrt{1+\mathbb{B}^2}\,dy=dA\right]$$
one gets
\begin{align}
\int_{-\infty}^{\mathbb{A}}-\frac{1}{\sqrt{1+\mathbb{B}^2}}\phi\left(\frac{A}{\sqrt{1+\mathbb{B}^2}}\right)\,dA=-\frac{1}{\sqrt{1+\mathbb{B}^2}}\sqrt{1+\mathbb{B}^2}\int_{-\infty}^{\mathbb{A}/\sqrt{1+\mathbb{B}^2}}\phi(y)\,dy=-\Phi({\mathbb{A}/\sqrt{1+\mathbb{B}^2}}).
\end{align}
The problem here is that this number should obviously be positive, so at some point I am missing a signal. As the computations seem sound to me, I would like to see if anyone could help me to find my mistake. 
Many thanks to you all.
 A: The horror, the horror... :-)
Recall that $\Phi(x)=P[X\leqslant x]$ for every $x$, where the random variable $X$ is standard normal, and that, for every suitable function $u$,
$$
\int_{-\infty}^{+\infty}u(x)\phi(x)\mathrm dx=E[u(Y)],
$$
where the random variable $Y$ is standard normal. Using this for $u=\Phi$, one sees that the integral $I$ you are interested in is
$$
I=P[X\leqslant B^{-1}(Y-A))]=P[BX\geqslant Y-A]=P[Z\leqslant A],
$$
where $Z=Y-BX$, where $X$ and $Y$ are i.i.d. standard normal, and where we used the fact that $B\lt0$ to reverse the inequality sign. Now, the random variable $Z$ is centered gaussian with variance $\sigma^2=1+B^2$, hence $Z=\sigma U$ with $U$ standard normal, and
$$
I=P[U\leqslant A/\sigma]=\Phi(A/\sqrt{1+B^2}).
$$
A: When you do the change of variable, you are doing:
\begin{align}
\left[y\longmapsto f\frac{\sqrt{1+B^2}}{B}-\frac{A}{B\sqrt{1+B^2}}\Longrightarrow df=\frac{B}{\sqrt{1+B^2}}\,dy\right]
\end{align}
.. which is correct. However, we need to apply the same change of variable to the bounds, ie to $-\infty$ and $+\infty$. Given that $\mathbb{B}$ is a negative constant, the bounds will change sign, becoming $+\infty$ and $-\infty$. So we will have:
\begin{align}
\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty\exp\left(-\frac{1}{2B^2}\left[f\sqrt{1+B^2}-\frac{A}{\sqrt{1+B^2}}\right]^2\right)\,df=\frac{B}{\sqrt{1+B^2}}\int_{\infty}^{-\infty}\phi(y)\,dy=-\frac{B}{\sqrt{1+B^2}}
\end{align}
This sign change will then propagate through to the final answer, reversing its sign too.
Full working, based off your working above, but with some buggettes removed:
Start with:
$$
I = 
\def\A{\mathbb{A}}
\def\B{\mathbb{B}}
\int_{-\infty}^\infty \Phi\left(
   \frac{f - \A}
       {\B}
   \right)\phi(f)\, df
$$
Take derivative wrt $\A$:
$$
\partial_\A I = \int_{-\infty}^\infty
\partial_\A \left(
    \Phi \left(
        \frac{f - \A}{\B}
    \right)
    \phi(f)
\right)
\, df
$$
$$
=\int_{-\infty}^\infty
\left(
   \frac{-1}{\B}
\right)
\phi\left(
   \frac{f - \A}{\B}
\right)
\phi(f)
\,
df
$$
Looking at $E_1 = \phi\left(\frac{f - \A}{\B}\right) \phi(f)$:
$$
E_1 = \frac{1}{2\pi}
\exp \left(
    - \frac{1}{2}
    \left(
        \frac{f^2 - 2f\A + \A^2 + \B^2f^2}
           {B^2}
    \right)
\right)
$$
$$
=\frac{1}{2\pi} \exp\left( - \frac{1}{2\B^2}
   \left(
   \left(
   \sqrt{(1 + \B^2)}f - \A\frac{1}{\sqrt{1 + \B^2}}
   \right)^2
   - \frac{\A^2}
      {1 + \B^2}
    + \A^2
\right)
\right)
$$
$$
- \frac{\A^2}{1+\B^2} + \A^2
$$
$$
= \frac{-\A^2 + \A^2 + \A^2\B^2}
  {1 + \B^2}
  $$
$$
= \frac{\A^2\B^2}
   {1 + \B^2}
$$
Therefore $E_1$ is:
$$
\frac{1}{2\pi}
\exp \left(
   -\frac{1}{2\B^2}
   \frac{\A^2\B^2}{1 + \B^2}
\right)
\exp \left(
    - \frac{1}{2\B^2} \left(
       f\sqrt{1 + \B^2} - \frac{\A}{\sqrt{1 + \B^2}}
    \right)^2
\right)
$$
$$
=
\frac{1}{2\pi}
\exp \left(
   -\frac{\A^2}{2(1 + \B^2)}
\right)
\exp \left(
    - \frac{1}{2\B^2} \left(
       f\sqrt{1 + \B^2} - \frac{\A}{\sqrt{1 + \B^2}}
    \right)^2
\right)
$$
Make change of variable:
$$
y = f\frac{\sqrt{1 + \B^2}}{\B} - \frac{A}{\B\sqrt{1 + \B^2}}
$$
Therefore:
$$
dy = \frac{\sqrt{1 + \B^2}}{\B}\,df
$$
For the limits, we have $f_1 = -\infty$, and $f_2 = \infty$
$\sqrt{1 + \B^2}$ is always positive. $\B$ is always negative. Therefore:
$$
y_1 = +\infty, y_2 = -\infty
$$
Therefore:
$$
\partial_\A I =
\frac{-1}{\B}\int_{+\infty}^{-\infty} \frac{1}{2\pi}
\exp \left(
  - \frac{\A^2} {2(1+\B^2)}
\right)
\exp \left(
  - \frac{1}{2} y^2
\right)
\frac{\B}{\sqrt{1 + \B^2}}
\, 
dy
$$
$$
= 
\frac{1}{\sqrt{2\pi}}
\frac{1}{\B}
\frac{\B}{\sqrt{1 + \B^2}}
\exp \left(
  - \frac{\A^2}{2(1 + \B^2)}
\right)
\int_{-\infty}^\infty
\frac{1}{\sqrt{2\pi}}
\exp\left(
  -\frac{1}{2} y^2
\right)
\,
dy
$$
$$
= \frac{1}{\sqrt{2\pi}}
\frac{1}{\sqrt{1+\B^2}}
\exp \left(
   - \frac{\A^2}{2(1 + \B^2)}
\right)
(1)
$$
$$
= \frac{1}{\sqrt{1 + \B^2}}
\phi\left(
   \frac{\A}{\sqrt{1 + \B^2}}
\right)
$$
Now we need to re-integrate back up again, since we currently have $\partial_\A I$, and we need $I$.
Since we dont have limits, we'll need to find at least one known point.
We have the following integral:
$$
I = \frac{1}{\sqrt{1 + \B^2}} \int \phi\left(
    \frac{\A}{\sqrt{1 + \B^2}}
\right)
\,d\A
$$
$$
= \frac{1}{\sqrt{1 + \B^2}}
\sqrt{1 + \B^2}
\Phi\left(
   \frac{\A}
      {\sqrt{1 + \B^2}}
\right)
+ C
$$
... where $C$ is a constant of integration
$$
= \Phi \left(
   \frac{\A}{\sqrt{1 + \B^2}}
\right)
+ C
$$
Looking at the original integral, we had/have:
$$
I = \int_{-\infty}^\infty \Phi\left(
   \frac{f - \A}{\B}
\right)
\phi(f)
\,
df
$$
We can see that, given that $\B$ is negative, as $\A \rightarrow \infty$, $\Phi\left(\frac{f-\A}{\B}\right) \rightarrow \Phi(\infty) = 1$.
Therefore, as $\A \rightarrow \infty$, $\int_{-\infty}^\infty \Phi(\cdot)\phi(f)\,df \rightarrow 1$
Meanwhile, looking at the later expression for $I$, ie:
$$
I= \Phi\left(
   \frac{\A}{\sqrt{1 + \B^2}}
\right) + C
$$
... as $\A \rightarrow +\infty$, $\Phi\left( \frac{\A}{\sqrt{1 + \B^2}} \right) \rightarrow 1$
But we know that as $\A \rightarrow +\infty$, $I \rightarrow 1$.
Therefore, $C = 0$
Therefore:
$$
I = \Phi \left(
   \frac{\A}
      {\sqrt{1 + \B^2}}
\right)
$$
