How to find this integral... $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\left(3x^2+2\sqrt2xy+3y^2\right)}dxdy$$
I have no idea how to integrate this function. If the middle $xy$ term would not have been present it would have been easy. But the $xy$ term is causing a problem. 
 A: The quadratic form $3x^2+2\sqrt 2 xy + 3y^2$ can be "diagonalised" by the (orthogonal) change of variables $x=(u+v)/\sqrt 2$ and $y=(u-v)/\sqrt 2$. They give 
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \mathrm e^{-[(3+\sqrt 2)u^2 + (3-\sqrt 2)v^2]}~\mathrm du~\mathrm dv$$
Next, take the change of coordinates $u=r\sqrt{3-\sqrt 2}\,\cos\theta$ and $v=r\sqrt{3+\sqrt 2}\,\sin\theta$. This gives $$(3+\sqrt 2)u^2 + (3-\sqrt 2)v^2 \leadsto 7r^2$$
where $-\infty < x,y < \infty$ becomes $0 < r< \infty$ and $0<\theta <2\pi$.
Moreover, $\mathrm du \wedge \mathrm dv = r\sqrt 7~\mathrm dr \wedge \mathrm d\theta.$ Hence
\begin{eqnarray*}
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \mathrm e^{-[(3+\sqrt 2)u^2 + (3-\sqrt 2)v^2]}~\mathrm du~\mathrm dv &=& \sqrt 7\int_0^{2\pi}\int_0^{\infty} r\mathrm e^{-7r^2}~\mathrm dr~\mathrm d\theta \\ \\
&=& \sqrt 7\int_0^{2\pi} \left[ -\frac{1}{14}e^{7r^2}\right]_{0}^{\infty}~\mathrm d\theta \\ \\
&=& \sqrt 7\int_0^{2\pi} \frac{1}{14}~\mathrm d\theta \\ \\
&=& \frac{\sqrt 7}{7}\pi
\end{eqnarray*}
A: $$I=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-(3x^2+2\sqrt2xy+3y^2)}dxdy$$
Let $x=r\cos\theta$, $y=r\sin\theta$.
Converting to polar form,
$$\begin{align}I&=\int_{0}^{2\pi}\int_{0}^{\infty}e^{-(3r^2\cos^2\theta+2\sqrt2r^2\sin\theta\cos\theta+3r^2\sin^2\theta)}rdrd\theta\\
&=\int_{0}^{2\pi}\int_{0}^{\infty}e^{-r^2(3+2\sqrt2\sin\theta\cos\theta)}rdrd\theta\\
&=\frac12\int_{0}^{2\pi}\int_{0}^{\infty}e^{-u(3+2\sqrt2\sin\theta\cos\theta)}dud\theta\\
&=\frac12\int_{0}^{2\pi}-\frac1{(3+2\sqrt2\sin\theta\cos\theta)}\left[e^{-u(3+2\sqrt2\sin\theta\cos\theta)}\right]^{\infty}_0d\theta\\
&=\frac12\int_0^{2\pi}\frac1{3+2\sqrt2\sin\theta\cos\theta}d\theta\\
&=\frac12\int_0^{2\pi}\frac1{3+\sqrt2\sin2\theta}d\theta\\
&=\frac14\int_0^{4\pi}\frac1{3+\sqrt2\sin v}dv\\
&=\frac12\int_0^{2\pi}\frac1{3+\sqrt2\sin v}dv\\
&=\frac12\frac{2\pi}{\sqrt{3^2-2}}\\
&=\frac{\pi}{\sqrt7}\end{align}$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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With $\ds{x \equiv u + v}$ and $\ds{y = u - v}$: 
\begin{align}
&\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp\pars{-\bracks{3x^{2} + 2\root{2}xy + 3y^{2}}}\,\dd x\,\dd y
\\[5mm] = &\
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\exp\pars{-\bracks{6 + 2\root{2}}u^{2}}\exp\pars{-\bracks{6 - 2\root{2}}v^{2}}\,
\
\overbrace{\verts{\partial\pars{x,y} \over \partial\pars{u,v}}}^{\ds{2}}\,\dd u\,\dd v
\\[5mm] = &\
\bbx{\ds{\root{7} \over 7}\,\pi} \approx 1.1874
\end{align}
