Evaluate $\int_0^{\infty}\int_0^{\infty}e^{-x^2-2xy-y^2}\,dx\,dy$

Question

I would like to compute the following, $$ \int_0^{\infty}\int_0^{\infty}e^{-x^2-2xy-y^2}\ dx\,dy $$ It is obvious that we can rewrite the integral above to, $$ \int_0^{\infty}\int_0^{\infty}e^{-(x+y)^2}\ dx\,dy $$ so we are ending up with something looking like a gaussian integral. I think that a smart substitution would help but all I tried ended up to be something I am not able to compute...

I really would appreciate any hint.

Thanks in advance!

Answer 1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}\int_{0}^{\infty}\expo{-\pars{x + y}^{2}}\,\dd y\,\dd x:\ {\large ?}}$

\begin{align} &\color{#66f}{\large% \int_{0}^{\infty}\int_{0}^{\infty}\expo{-\pars{x + y}^{2}}\,\dd y\,\dd x} \\[5mm] = &\ \int_{0}^{\infty}\int_{x}^{\infty}\expo{-y^{2}}\,\dd y\,\dd x =\left.\int_{0}^{\infty}\int_{0}^{\infty}\expo{-y^{2}}\,\dd y\,\dd x \right\vert_{y\ >\ x} \\[3mm]&=\left.\int_{0}^{\infty}\expo{-y^{2}}\int_{0}^{\infty}\,\dd x\,\dd y \right\vert_{x\ <\ y} = \int_{0}^{\infty}\expo{-y^{2}}\int_{0}^{y}\,\dd x\,\dd y =\int_{0}^{\infty}y\expo{-y^{2}}\,\dd y \\[5mm] = &\ \left.-\,\half\,\expo{-y^{2}}\right\vert_{0}^{\infty} = \color{#66f}{\Large\half} \end{align}

Answer 2

Switch to polar coordinates i.e $x=r\cos\theta$, $y=r\sin\theta$ and $dx\,dy=r\,dr\,d\theta$ to obtain: $$\int_0^{\pi/2} \int_0^{\infty} re^{-r^2(1+\sin(2\theta))}dr\,d\theta=\int_0^{\pi/2} \frac{1}{2(1+\sin(2\theta))}\,d\theta$$ Write $\sin(2\theta)=\frac{2\tan\theta}{1+\tan^2\theta}$ to get: $$\int_0^{\pi/2} \frac{1}{2(1+\sin(2\theta))}\,d\theta=\frac{1}{2}\int_0^{\pi/2} \frac{\sec^2\theta}{1+\tan^2\theta+2\tan\theta}\,d\theta$$ Use the substitution $\tan\theta=t \Rightarrow \sec^2\theta\,d\theta=dt$: $$\frac{1}{2}\int_0^{\pi/2} \frac{\sec^2\theta}{1+\tan^2\theta+2\tan\theta}\,d\theta=\frac{1}{2}\int_0^{\infty} \frac{dt}{(1+t)^2}=\boxed{\dfrac{1}{2}}$$

Answer 3

Theorem : $$ \iint_A f(x,y)\ dx\,dy=\iint_B g(u,v) |J|\ du\,dv, $$ where $J$ is Jacobian.

Now, using parametric equations $u=x+y$ and $v=x$ then its Jacobian is $-1$. The corresponding regions are $0<x<\infty\;\Rightarrow\;0<v<\infty$ and $0<y<\infty\;\Rightarrow\;0<u-v<\infty$. Hence \begin{align} \int_0^{\infty}\int_0^{\infty}e^{\large-(x+y)^2}\ dx\,dy&=\int_{v=0}^\infty\int_{u=v}^\infty e^{\large-u^2}\ du\,dv\\ &=\int_{u=0}^\infty\int_{v=0}^u e^{\large-u^2}\ dv\,du\\ &=\int_{u=0}^\infty u\ e^{\large-u^2}\ du\qquad;\qquad\text{let}\ t=u^2\;\Rightarrow\;dt=2u\ du\\ &=\frac12\int_{t=0}^\infty \ e^{\large-t}\ dt\\ &=\large\color{blue}{\frac12}. \end{align}