The larger of two normals

Idea. Write the maximum as the average plus half the gap. The average has mean zero, and the gap is the absolute value of a normal variable, whose mean is known.

Step 1: decompose. $$\max (X,Y)=\frac{X+Y}{2}+\frac{|X-Y|}{2}.$$ The first term has mean $0$.

Step 2: the gap. $X-Y\sim \mathcal{N}(0,2)$. For $Z\sim \mathcal{N}(0,{\sigma }^2)$, $$\mathbb{E}|Z|=2{\int }_0^{\infty }\frac{z}{\sigma \sqrt{2\pi }}{e}^{-z^2/(2{\sigma }^2)}dz=\sigma \sqrt{\frac{2}{\pi }}.$$ With $\sigma =\sqrt{2}$, $\mathbb{E}|X-Y|=\frac{2}{\sqrt{\pi }}$.

Step 3. $$\mathbb{E}[\max (X,Y)]=\frac{1}{2}\cdot \frac{2}{\sqrt{\pi }}=\frac{1}{\sqrt{\pi }}\approx \mathrm{0.564.}$$

Going further. The same trick gives $\mathbb{E}[\min (X,Y)]=-\frac{1}{\sqrt{\pi }}$. For two normals with correlation $\rho$, the gap has variance $2(1-\rho )$ and the answer becomes $\sqrt{(1-\rho )/\pi }$: the more correlated the two, the less the maximum gains.