calculus - How to Evaluate $int^infty_0int^infty_0e^{-(x+y)^2} dx dy$

How do you get from $$\int^\infty_0\int^\infty_0e^{-(x+y)^2} dx\ dy$$ to

$$\frac{1}{2}\int^\infty_0\int^u_{-u}e^{-u^2} dv\ du?$$ I have tried using a change of variables formula but to no avail.
Edit: Ok as suggested I set $u=x+y$ and $v=x-y$, so I can see this gives $dx dy=\frac{1}{2}dudv$ but I still can't see how to get the new integration limits. Sorry if I'm being slow.