Help evaluating integral (anything simple that I am missing?)

Change the variables (rotation by $\pi/4$) $$ \begin{cases} u&=&\frac{1}{\sqrt2}(x-y),\\ v&=&\frac{1}{\sqrt2}(x+y) \end{cases} $$ to get $$ \frac{1}{2\pi}\int_{-\infty}^{+\infty}\int_{-\sqrt2}^{\sqrt2}e^{-\frac{u^2+v^2}{2}}\,dvdu=\frac{1}{2\pi}\int_{-\infty}^{+\infty}e^{-u^2/2}\,du\int_{-\sqrt2}^{\sqrt2}e^{-v^2/2}\,dv=\frac{1}{\sqrt\pi}\int_{-1}^1e^{-s^2}\,ds=\color{red}{\operatorname{erf}(1)}. $$

The integral is $\ \mathbb{P}\left(-2\le X+Y\le 2\right)\ $, where $\ X,Y\ $ are independent standard normal variates. In these circumstances $\ \frac{X+Y}{\sqrt {2}}\ $ is also a standard normal variate, so the value of the integral is $\ \mathcal{N}_{(0,1)}(\sqrt{2})-\mathcal{N}_{(0,1)}(-\sqrt{2})\approx 0.8427 $.