How to Evaluate $\int^\infty_0\int^\infty_0e^{-(x+y)^2} dx\ dy$

Question

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.

Answer 1

Hint: Try $u = x+y$, $v = x-y$

Answer 2

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 $$ \frac{\partial (x,y)}{\partial(u,v)}=\frac{1}{2} $$

Answer 3

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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}