Integral with Normal Distributions I know that the following equality is true for any $a$ and $\sigma$ (I have solved it numerically):
$$\int\nolimits_{-\infty}^{+\infty}\Phi\left(\frac{a-x}{\sigma}\right)\frac1{\sigma} \phi\left(\frac{x-a}{\sigma}\right)\mathrm dx=\frac12$$
Where $\Phi$ and $\phi$ respectively denote the CDF and PDF of the standard normal distribution, $a$ is any given real number, and $\sigma$ is a positive real number.
I have tried for a while but I am unable to prove it. Can anyone help me?
 A: NB: Throughout what follows, we denote the standard normal density function by $\varphi(x) = (2\pi)^{-1/2} \exp(-\frac{1}{2} x^2)$ and its cumulative distribution function by $\Phi(x)$.

Method 1: Avoid calculus.
This is a consequence of a much more general result and, in fact, has nothing in particular to do with the normal distribution at all. Here is a slightly simplified version.
Let $F$ be a strictly increasing distribution function on some interval $(a,b)$ such that $F(a) = 0$ and $F(b) = 1$. We allow $a = -\infty$ and $b=\infty$, so that we can handle the case where $F$ is defined on the entire real line, as in your example.
Suppose $X$ is a random variable with distribution function $F$. Then, $Y = F(X)$ has a uniform distribution on $(0,1)$. The proof is simple. For $y \in (0,1)$,
$$
\renewcommand{\Pr}{\mathbb{P}}
\Pr(Y \leq y) = \Pr( F(X) \leq y) = \Pr(X \leq F^{-1}(y)) = F( F^{-1}(y) ) = y \> ,
$$
where the inverse $F^{-1}$ exists by the hypothesis that $F$ is strictly increasing on $(a,b)$.
Hence, $Y$ is distributed uniformly on $(0,1)$ and, as a consequence, $\newcommand{\e}{\mathbb{E}}\e Y = 1/2$.
Note that for your particular case, you start with $X \sim \mathcal{N}(a, \sigma^2)$ and so $Y = \Phi((X-a)/\sigma)$. Your problem can easily be seen to be equivalent to asking for $\e (1 - Y) = \e Y = 1/2$.

Method 2: Hammer and tongs (i.e., use calculus).
Note that your integral can be written as
$$
\int_{-\infty}^\infty \int_{-\infty}^{- (x-a)/\sigma} \varphi(y) \frac{1}{\sigma} \varphi((x-a)/\sigma) \newcommand{\rd}{\mathrm{d}} \,\rd y \, \rd x = \int_{-\infty}^\infty \int_{-\infty}^{\sigma y + a} \varphi(y) \frac{1}{\sigma} \varphi((x-a)/\sigma) \,\rd x \, \rd y \> .
$$
Now, change variables using $u = (x-a)/\sigma$, which gives the integral
$$
\int_{-\infty}^\infty \int_{-\infty}^y \varphi(u) \varphi(y) \, \rd u \,\rd y \>.
$$
Exchanging the order of the iterated integrals, we get
$$
\int_{-\infty}^\infty \int_{-\infty}^y \varphi(u) \varphi(y) \, \rd u \,\rd y = \int_{-\infty}^\infty \int_u^\infty \varphi(u) \varphi(y) \, \rd y \,\rd u \>.
$$
But, since $\varphi$ is a probability density function
$$
\int_{-\infty}^\infty \int_{-\infty}^y \varphi(u) \varphi(y) \, \rd u \,\rd y + \int_{-\infty}^\infty \int_y^\infty \varphi(u) \varphi(y) \, \rd u \,\rd y = 1 \>,
$$
and so, by symmetry, the integral must be $1/2$.
(All exchanges of order of integration are valid by Fubini's theorem.)
A: cardinal already posted a solution, but I'll include mine as an alternative.
Being more accustomed to the error function $\mathrm{erf}$ than with $\Phi$, I'll use the formulae
$$\phi(x)=\frac1{\sqrt{2\pi}}\exp\left(-\frac{x^2}{2}\right)$$
$$\Phi(x)=\frac12\left(1+\mathrm{erf}\left(\frac{x}{\sqrt{2}}\right)\right)$$
and consider the integral
$$\frac1{2\sigma\sqrt{2\pi}}\int_{-\infty}^{\infty}\exp\left(-\frac{(x-a)^2}{2\sigma^2}\right)\left(1-\mathrm{erf}\left(\frac{x-a}{\sqrt{2}\sigma}\right)\right) \mathrm dx$$
Letting $u=\frac{x-a}{\sqrt{2}\sigma}$, $\mathrm du=\frac{\mathrm dx}{{\sqrt{2}\sigma}}$, the integral becomes
$$\frac1{2\sqrt{\pi}}\int_{-\infty}^{\infty}\exp\left(-u^2\right)\left(1-\mathrm{erf}\left(u\right)\right) \mathrm du$$
and then we use the fact that
$$\int_{-\infty}^{\infty}f(u)\mathrm du=\frac12\int_{-\infty}^{\infty}(f(u)+f(-u)\mathrm du$$
and that $\mathrm{erf}$ is an odd function ($\mathrm{erf}(-x)=-\mathrm{erf}(x)$) to arrive at the integral
$$\frac1{2\sqrt{\pi}}\int_{-\infty}^{\infty}\exp\left(-u^2\right)\mathrm du$$
whose evaluation you should be familiar with...
A: This follows from a basic change of variables, for which one just needs to know that $\Phi'=\phi$ and what the limits of $\Phi$ at $\pm\infty$ are, but certainly not the precise form of $\Phi$. 
To wit, consider $z=\displaystyle\Phi\left(\frac{x-a}{\sigma}\right)$, then $\mathrm{d}z=\displaystyle\phi\left(\frac{x-a}{\sigma}\right)\frac{\mathrm{d}x}{\sigma}=\displaystyle\phi\left(\frac{a-x}{\sigma}\right)\frac{\mathrm{d}x}{\sigma}$ because $\Phi'=\phi$ (first equality) and because $\phi$ is even (second equality). 
The bounds of the integral in $z$ are $\Phi(-\infty)=0$ for $x=-\infty$ and $\Phi(+\infty)=1$ for $x=+\infty$, hence the integral is $\displaystyle\int_0^1z\mathrm{d}z=\left[\frac{z^2}2\right]_0^1=\frac12$.
