Expected values of $\max(X,Y)$ and $\min(X,Y)$ for $N(\mu,\sigma^2)$ distributed $X$ and $Y$ Suppose that $X$ and $Y$ are independent and  $N(\mu,\sigma^2)$  distributed. Then 
$E(\min(X,Y))=\mu-\frac{\sigma}{\sqrt{\pi}}$  and  $E(\max(X,Y))=\mu+\frac{\sigma}{\sqrt{\pi}}$. 
I tried to verify these results using the relations  $E(\max(X,Y))=\mu+\sigma\int_{-\infty}^{\infty}t \frac{d}{dt}\Phi(t)^2dt$ and $\max(X,Y)+\min(X,Y)=X+Y$, but I was not able to evaluate the integral.
 A: $$\begin{eqnarray*}\mathbb{E}[\min(X,Y)]&=&\frac{1}{2\pi\sigma^2}\iint_{\mathbb{R}^2}\min(x,y)\,e^{-\frac{(x-\mu)^2}{2\sigma^2}}e^{-\frac{(y-\mu)^2}{2\sigma^2}}\\&=&\frac{1}{2\pi\sigma^2}\iint_{\mathbb{R}^2}\min(x+\mu,y+\mu)\,\exp\left(-\frac{x^2+y^2}{2\sigma^2}\right)\,dx\,dy\\&=&\mu+\frac{1}{2\pi\sigma^2}\iint_{\mathbb{R}^2}\min(x,y)\,\exp\left(-\frac{x^2+y^2}{2\sigma^2}\right)\,dx\,dy\\&=&\mu+\frac{1}{\pi\sigma^2}\iint_{y\leq x}y\,\exp\left(-\frac{x^2+y^2}{2\sigma^2}\right)\,dx\,dy\\&=&\mu+\frac{1}{\pi \sigma^2}\int_{-\frac{3\pi}{4}}^{\frac{\pi}{4}}\int_{0}^{+\infty}\rho^2\sin(\theta) \,e^{-\frac{\rho^2}{2\sigma^2}}\,d\rho\,d\theta\\&=&\mu-\frac{\sqrt{2}}{\pi \sigma^2}\int_{0}^{+\infty}\rho^2\,e^{-\frac{\rho^2}{2\sigma^2}}\,d\rho\\&=&\mu-\frac{\sqrt{2}}{\pi \sigma^2}\cdot \sigma^3\sqrt{\frac{\pi}{2}}=\color{red}{\mu-\frac{\sigma}{\sqrt{\pi}}}\tag{1}\end{eqnarray*}$$
and since $\min(X,Y)+\max(X,Y)=X+Y$ it follows that
$$ \mathbb{E}[\max(X,Y)] = 2\mu-\mathbb{E}[\min(X,Y)]=\color{red}{\mu+\frac{\sigma}{\sqrt{\pi}}} \tag{2}$$
by the linearity of expectation. You may also exploit the identity:
$$ \max(X,Y)-\min(X,Y) = \left|X-Y\right| \tag{3}$$
and the fact that $X-Y$ is normally distributed with variance $2\sigma^2$ and mean zero, so:
$$ \mathbb{E}[\left|X-Y\right|]=\frac{1}{\sqrt{2}\,\sigma\sqrt{2\pi}}\int_{0}^{+\infty}2x \exp\left(-\frac{x^2}{4\sigma^2}\right)\,dx=\frac{2\sigma}{\sqrt{\pi}}\tag{4}$$
that together with $\mathbb{E}[\min(X,Y)]+\mathbb{E}[\max(X,Y)]=2\mu$ is enough to recover both $(1)$ and $(2)$.
