Brutal gaussian integral of death $\int_{\mathbb{R}} x \Phi(x) \phi(Bx-b)$ Ciao,
I was making some computation and I've been stucked in this one.
Let $B$ and $b$ be positive contant. We call $\phi(x)$ standard gaussian distribution and $\Phi(x)$ its cumulative function, i.e.
$$
\phi(x) = \frac{1}{\sqrt{2 \pi}}e^{-\frac{x^2}{2}}
$$
$$
\Phi(x) = \int_{-\infty}^x \phi(s) ds
$$
then compute


$$
\int_{-\infty}^{+\infty}x\Phi(x)\phi(Bx - b) dx
$$


If it helps I can proove this result:
$$
\int_{-\infty}^{+\infty}x\Phi(x)\phi(x) dx = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} e^{-x^2}= \frac{1}{\sqrt{2}}
$$
Any suggestion or hint will be appreciated,
thank you!
Ciao
AM
 A: We have
$$ \int_{- \infty}^x ds \ \phi(s) = \int_{- \infty}^0 ds \ \phi(s + x) \ .$$
Using Fubini's theorem, we have
$$ \int_{-\infty}^\infty dx \ x \Phi(x) \phi(Bx - b) = \int_{- \infty}^0 ds \int_{-\infty}^\infty dx \ x \phi(s + x) \phi(Bx + b) \ .$$
Computing the inner integral, we have
$$ \int_{-\infty}^\infty dx \ x \phi(s + x) \phi(Bx + b) = \frac{(bB - s)e^{- \frac{(Bs + b)^2}{2(1 + B^2)}} }{\sqrt{2 \pi}(1 + B^2)^\frac{3}{2}} \ .$$
We have
$$ \int_{- \infty}^0 ds \int_{-\infty}^\infty dx \ x \phi(s + x) \phi(Bx + b) = \int_{0}^\infty  ds \ \frac{(bB + s)e^{- \frac{(- Bs + b)^2}{2(1 + B^2)}} }{\sqrt{2 \pi}(1 + B^2)^\frac{3}{2}} \ .$$
From here, I suppose one has a representation in terms of the error function. I do not see any other way to simplify this expression.
If we set $B = 1$ and $b = 0$, then the value of this integral agrees with the value that Anne calculated in the comments.
A: We have:
\begin{eqnarray}
I:= \int\limits_{{\mathbb R}} x \Phi(x) \phi(B x-b) dx = 
\int\limits_{{\mathbb R}} (\frac{y+b}{B}) \Phi(\frac{y+b}{B}) \phi(y) \frac{dy}{B}
\end{eqnarray}
Now since $\phi(y)^{'} = -y \phi(y)$ we can treat one of the resulting integrals with integration by parts. We have:
\begin{eqnarray}
\int\limits_{{\mathbb R}} \underbrace{y \phi(y)}_{-\phi^{'}(y)} \Phi(\frac{y+b}{B}) dy=\int\limits_{\mathbb R} \phi(y) \phi(\frac{y+b}{B}) \frac{1}{B} dy = \phi(\frac{b}{\sqrt{1+B^2}}) \frac{1}{\sqrt{1+B^2}}
\end{eqnarray}
where the last step is done by completing the integrand to square.
The second integral is treated by differentiating it with respect to the $b$ parameter. We have:
\begin{eqnarray}
f(b)&:=& \int\limits_{\mathbb R} \phi(y) \Phi(\frac{y+b}{B})dy \quad \Rightarrow \quad f^{'}(b) = \int\limits_{\mathbb R} \phi(y) \phi(\frac{y+b}{B}) \frac{1}{B} dy = \phi(\frac{b}{\sqrt{1+B^2}}) \frac{1}{\sqrt{1+B^2}}\\
&\Rightarrow & f(b) = \Phi(\frac{b}{\sqrt{1+B^2}})
\end{eqnarray}
where we used the boundary condition $f(-\infty)=0$. Bringing all together we have:
\begin{equation}
I = \frac{1}{B^2} \left[\phi(\frac{b}{\sqrt{1+B^2}}) \frac{1}{\sqrt{1+B^2}}+b \Phi(\frac{b}{\sqrt{1+B^2}}) \right]
\end{equation}
