Evaluate $\int_0^{\infty}\int_0^{\infty}e^{-x^2-2xy-y^2}\,dx\,dy$ 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!
 A: 
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}
A: $\newcommand{\+}{^{\dagger}}
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$\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}

A: 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 $\tan\theta=t \Rightarrow \sec^2\theta\,d\theta=dt$:
$$\frac{1}{2}\int_0^{\pi/2} \frac{\sec^2\theta}{1+\tan^2\theta+2\tan\theta}\,d\theta=\frac{1}{2}\int_0^{\infty} \frac{dt}{(1+t)^2}=\boxed{\dfrac{1}{2}}$$
