$\int^{\infty}_{-\infty} \int^{\infty}_{-\infty} e^{-\left(2x^2+2xy+2y^2\right)} dx\,dy\,$

I need to evaluate $$\displaystyle\int^{\infty}_{-\infty} \int^{\infty}_{-\infty} e^{-\left(2x^2+2xy+2y^2\right)} dx\,dy\,$$

I think I'll need $\displaystyle\int^{\infty}_{-\infty} e^{-x^2} dx\,=\sqrt{\pi}$ . But I'm not able to apply it properly.

Solution 1:

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\overbrace{\color{#66f}{\large% \int^{\infty}_{-\infty}\int^{\infty}_{-\infty} \exp\pars{-\bracks{2x^{2} + 2xy + 2y^{2}}}\,\dd x\,\dd y}} ^{\ds{\dsc{x}=\dsc{\mu + \nu}}\,,\ \dsc{y}=\dsc{\mu - \nu}} \\[5mm]&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \exp\pars{-2\bracks{3\mu^{2} + \nu^{2}}}\,\ \overbrace{% \verts{\partiald{\pars{x,y}}{\pars{\mu,\nu}}}}^{\ds{=\ \dsc{2}}}\,\dd\mu\,\dd\nu \\[5mm]&=2\int_{-\infty}^{\infty}\exp\pars{-6\mu^{2}}\,\dd\mu \int_{-\infty}^{\infty}\exp\pars{-2\nu^{2}}\,\dd\nu \\[5mm]&=2 \bracks{{1 \over \root{6}}\ \overbrace{\int_{-\infty}^{\infty}\exp\pars{-\mu^{2}}\,\dd\mu}^{\dsc{\root{\pi}}}} \bracks{{1 \over \root{2}}\ \overbrace{\int_{-\infty}^{\infty}\exp\pars{-\nu^{2}}\,\dd\mu}^{\dsc{\root{\pi}}}} ={\pi \over \root{3}}=\color{#66f}{\large{\root{3} \over 3}\,\pi} \end{align}

Solution 2:

OK, the problem seems to be completing the square. What one should aim for is to collect the mixed term $xy$ into a squared term. Here I do it with the $x^2$-term: $$ 2x^2+2xy+2y^2=2\bigl(x+\tfrac{1}{2}y\bigr)^2+\frac{3}{2}y^2. $$ Once that is done, you can change variables (for example $u=\sqrt{2}(x+y/2)$ and $v=\sqrt{3/2}y$. Can you proceed from here?