Prove $\int_{-\infty}^\infty \phi(x )\Phi(x - T)dx = 1 - \Phi(T/\sqrt{2})$. When I compute the conditional density $f(x)$ of $X_1$ given $X_1 + X_2 > T$, where $X_1, X_2$ are independently and identically distributed standard normal random variables, $T$ is any real number, I obtained
\begin{equation}
f(x) = \frac{1}{1 - \Phi(T/\sqrt{2})}\phi(x)\Phi(x - T),
\end{equation}
where $\phi(x)$ and $\Phi(x)$ are density and distribution functions of $N(0, 1)$.  This implies the integral identity
\begin{equation*}
\int_{-\infty}^\infty \phi(x)\Phi(x - T)\,dx = 1 - \Phi(T/\sqrt{2}).  \tag{$*$}
\end{equation*}
However, except for the special case $T = 0$, I can't verify $(*)$ analytically.  Is there a simple way to prove it?  If yes, can the same approach be applied to evaluate
$$\int_{-\infty}^\infty x\phi(x)\Phi(x - T)\,dx?$$

Let $f(y)=\int_{-\infty}^\infty\phi(x)\Phi(x-y)\,dx$, then $f(y)=1-\Phi(y/\sqrt2)$ is deduced easily from $$-f'(y)=\int_{-\infty}^\infty\phi(x)\phi(x-y)\,dx=\frac1{2\pi}e^{-y^2/4}\int_{-\infty}^\infty e^{-(x-y/2)^2}dx=\frac1{\sqrt2}\phi\left(\frac{y}{\sqrt2}\right)$$ and $f(y)\to0$ as $y\to+\infty$. The same way (or simply using $x\phi(x)=-\phi'(x)$ and IBP), $$\int_{-\infty}^\infty x\phi(x)\Phi(x-y)\,dx=\frac1{\sqrt2}\phi\left(\frac{y}{\sqrt2}\right).$$
