Expected Value of Normal Random Variable times its CDF As usual, let $\Phi$ and $\varphi$ denote the cumulative density function and the density function of a standard normal random variable.
On the wiki page "List of integrals of Gaussian functions", I have found an expected value integral involving  a standard normal r.v. and its cdf,
$$I=\int_{-\infty}^{\infty}x\varphi(x)\Phi(a+bx)dx=\frac{b}{\sqrt{1+b^2}}\varphi\left(\frac{a}{\sqrt{1+b^2}}\right),$$
for which I do not know how to do the last step in my solution:
My ansatz is to introduce a parameter integral, $I:=I(a)$, and finding its derivative:
$$
\begin{align*}
\frac{\partial I}{\partial a}&=\int_{-\infty}^{\infty}x\varphi(x)\varphi(a+bx)dx\\
&=\int_{-\infty}^{\infty}x\frac{e^{-\frac{1}{2}\left(x^2(1+b^2)+2abx+a^2\right)}}{2\pi}dx\\
&=a\frac{e^{-\frac{1}{2}\frac{a^2}{1+b^2}}}{\sqrt{1+b^2}\sqrt{2\pi}}\\
&=a\frac{\varphi\left(\frac{a}{\sqrt{1+b^2}}\right)}{\sqrt{1+b^2}}
\end{align*}
$$
Integrating the derivative, we obtain:
$$
\begin{align}
I&=\int \frac{\partial I}{\partial a}da + C\\
&=\frac{b}{\sqrt{1+b^2}}\varphi\left(\frac{a}{\sqrt{1+b^2}}\right)+C,
\end{align}
$$
which equals the solution on the wiki page plus a constant term $C$. From here on, I do not know how to get rid of the integration constant, i.e. how to show that $C=0$.
I do know that for $b=0$ it holds that $I=0$. Is this be sufficient to pin down $C$ to zero? Or do I miss something completely?
 A: Apart from parameter integral, we can solve the problem from the perspective of probability theory. Notice that 
$$I = \int_{-\infty}^{+\infty}x\Phi(a+bx)\varphi(x)dx = \mathbb{E}[X\Phi(a+bX)]$$
Let's say X and Z are independent standard normal variables, we have
$$\Phi(a+bX) = \mathbb{P}(Z\leq a+bX|X) = \mathbb{E}[1_{\{Z\leq a+bX\}}|X]$$
, then
$$\mathbb{E}[X\Phi(a+bX)] = \mathbb{E}\left[X\cdot\mathbb{E}[1_{\{Z\leq a+bX\}}|X]\right]=\mathbb{E}\left[\mathbb{E}[X\cdot1_{\{Z\leq a+bX\}}|X]\right]$$
For any random variables $\xi$ and $\eta$, applying the law of total expectation, we have
$$\mathbb{E}\xi = \mathbb{E}\left[\mathbb{E[\xi|\eta]}\right]$$
Using the upper equality for $\xi=X\cdot1_{\{Z\leq a+bX\}}$ and $\eta=X$, we obtain that
$$\mathbb{E}[X\Phi(a+bX)]=\mathbb{E}[X\cdot1_{\{Z\leq a+bX\}}]$$
Next, all we need to do is to calculate the double integral, for convenience, let's say $a>0$ and $b>0$(details as follows)
\begin{split}
\mathbb{E}[X\cdot1_{\{Z\leq a+bX\}}]&=\int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}\int_{\frac{z-a}{b}}^{+\infty}\frac{1}{\sqrt{2\pi}}xe^{-\frac{x^2}{2}}dxdz\\
&=\int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}\cdot\frac{1}{\sqrt{2\pi}}e^{-\frac{(z-a)^2}{2b^2}}dz\\
&=\frac{b}{\sqrt{1+b^2}}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}\frac{a^2}{(1+b^2)}}\int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi}\cdot\frac{b}{\sqrt{1+b^2}}}e^{-\frac{1}{2}\frac{(z-\frac{a}{1+b^2})^2}{\frac{b^2}{1+b^2}}}dz\\
&=\frac{b}{\sqrt{1+b^2}}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}\frac{a^2}{(1+b^2)}}\\
&=\frac{b}{\sqrt{1+b^2}}\varphi(\frac{a}{\sqrt{1+b^2}})
\end{split}
Note that, for other values of a and b, the deduction above also applies.
