How to Evaluate $\int^\infty_0\int^\infty_0e^{-(x+y)^2} dx\ dy$ 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.
 A: Hint:   Try $u = x+y$, $v = x-y$
A: Hint. Set $u=x+y$, $v=x-y$.
Then 
$$
\{(x,y): x,y\ge 0\}=\{(u,v) : u>0, v\in(-u,u)\},
$$
and
$$
dx\,dy=\frac{1}{2}du\,dv,
$$
as
$$
\frac{\partial (x,y)}{\partial(u,v)}=\frac{1}{2}
$$
A: $\newcommand{\+}{^{\dagger}}%
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 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
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With $x \equiv \rho\cos\pars{\theta}$, $y \equiv \rho\sin\pars{\theta}$ where $\rho \geq 0$ and $0 \leq \theta < 2\pi$ we'll get
$\ds{{\partial\pars{x,y} \over \partial\pars{\rho,\theta}} = \rho}$ such that

\begin{align}
&\color{#0000ff}{\large%
\int_{0}^{\infty}\int_{0}^{\infty}\expo{-\pars{x + y}^{2}}\,\dd x\,\dd y}
=
\int_{0}^{\pi/2}\dd\theta\int_{0}^{\infty}\expo{-\rho^{2}\bracks{1 + \sin\pars{2\theta}}}\rho\,\dd\rho
\\[3mm]&=
\int_{0}^{\pi/2}\dd\theta\,\left.%
{-\expo{-\rho^{2}\bracks{1 + \sin\pars{2\theta}}} \over 2\bracks{1 + \sin\pars{2\theta}}}
\right\vert_{\rho = 0}^{\rho \to \infty}
=
\half\int_{0}^{\pi/2}
{\dd\theta \over 1 + \sin\pars{2\theta}}
=
{1 \over 4}\int_{0}^{\pi}
{\dd\theta \over 1 + \sin\pars{\theta}} = \color{#0000ff}{\large\half}
\end{align}

since
\begin{align}
&\color{#0000ff}{\large{1 \over 4}\int_{0}^{\pi}{\dd\theta \over 1 + \sin\pars{\theta}}}
=\half\int_{0}^{\pi/2}{\dd\theta \over 1 + \sin\pars{\theta}}
=\half\int_{0}^{\pi/2}{1 - \sin\pars{\theta} \over \cos^{2}\pars{\theta}}\,\dd\theta
\\[3mm]&=
\half\,\lim_{\theta \to \pars{\pi/2}^{-}}\bracks{%
{\sin\pars{\theta} - 1 \over \cos\pars{\theta}}} + \half = \color{#0000ff}{\large\half}
\end{align}
