Evaluate :$I=\int_{-\infty}^\infty \int_{-\infty}^\infty \mathrm e^{-(2x^2+2xy+2y^2)}dxdy$ Evaluate:$$I=\int_{-\infty}^\infty \int_{-\infty}^\infty  \mathrm e^{-(2x^2+2xy+2y^2)}dxdy$$
I am just clueless to solve it.your help will be appreciated.
 A: Note that $$\int_{-\infty}^{\infty} \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{1}{2}{\big(\frac{x-\mu}{\sigma}\big)}^2}=1; ~~\mu \in \mathbb{R}, ~\sigma>0 ~\cdots (\star)$$
Here the integrand is the probability density function (pdf) of the Normal distribution (as it is called by the statisticians), aka the Gaussian distribution (as it is called by the mathematicians). And by definition, for any pdf $f$, we have $\int_{\Omega}f(x)dx=1$, where $\Omega$ is the domain of $f$. Here $\Omega=\mathbb{R}$.
Now, 
\begin{align}
I&=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(2x^2+2xy+2y^2)} dx dy\\
&=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-2\big\{(x+\frac{y}{2})^2+\frac{3y^2}{4}\big\}} dx dy\\
&=\int_{-\infty}^{\infty} e^{-2\frac{3y^2}{4}}\Big\{ \int_{-\infty}^{\infty} e^{-2(x+\frac{y}{2})^2} dx\Big\}dy\\
&=\frac{1}{2}\sqrt{2\pi}\int_{-\infty}^{\infty} e^{-\frac{3y^2}{2}}\bigg\{ \int_{-\infty}^{\infty} \frac{1}{\frac{1}{2}\sqrt{2\pi}}e^{-\frac{1}{2}\bigg(\frac{x-(-\frac{y}{2})}{\frac{1}{2}}\bigg)^2} dx\bigg\}dy\\
&=\frac{\sqrt{\pi}}{2}\int_{-\infty}^{\infty} \big( e^{-\frac{3y^2}{2}} \times 1 \big)dy ~~\mbox{[Using $(\star)$]}\\
&=\sqrt{\frac{\pi}{2}} \sqrt{\frac{2\pi}{3}} \int_{-\infty}^{\infty} \frac{1}{\frac{1}{\sqrt{3}}\sqrt{2\pi}}e^{-\frac{1}{2}\bigg( \frac{y-0}{\frac{1}{\sqrt{3}}} \bigg)^2} dy\\
&=\sqrt{\frac{\pi}{2}} \sqrt{\frac{2\pi}{3}} \times 1 ~~\mbox{[Again, using $(\star)$]}=\frac{\pi}{\sqrt{3}}
\end{align}
A: Hint: Gaussian Integral and the observation $x^2+xy+y^2=(x+ \frac {y}{2})^2+(\frac {\sqrt{3}}{2}y)^2$ is precisely what you need.
A: $$
\begin{align}
\int_0^{2\pi}\int_0^\infty e^{-(2+\sin(2\theta))\,r^2}r\,\mathrm{d}r\,\mathrm{d}\theta
&=\frac12\int_0^{2\pi}\frac{\mathrm{d}\theta}{2+\sin(2\theta)}\\
&=\frac14\int_0^{4\pi}\frac{\mathrm{d}\theta}{2+\cos(\theta)}\\
&=\frac12\oint_{|z|=1}\frac1{2+\frac{z+1/z}2}\frac{\mathrm{d}z}{iz}\\
&=\frac1i\oint_{|z|=1}\frac{\mathrm{d}z}{z^2+4z+1}\\
&=2\pi\operatorname*{Res}_{z=-2+\sqrt3}\left(\frac1{z^2+4z+1}\right)\\
&=2\pi\left.\left(\frac1{2z+4}\right)\right|_{z=-2+\sqrt3}\\
&=\frac\pi{\sqrt3}
\end{align}
$$
A: I find it strange how other answers don't keep the symmetry. The symmetry between $x$ and $y$ openly suggests substitution $u=\frac{1}{\sqrt2}(x+y)$, $v=\frac{1}{\sqrt2}(x-y)$.
$$u^2=\frac12x^2+xy+\frac12y^2$$
$$v^2=\frac12x^2-xy+\frac12y^2$$
$$3u^2+v^2=2x^2+2xy+2y^2$$
As this substitution is just a $45^\circ$ rotation of the $xy$ plane, the Jacobian is $1$. The integral splits into a product of two Gaussian integrals, which have a well known value.
$$I=\int_{-\infty}^\infty e^{-3u^2}{\rm d}u \int_{-\infty}^\infty e^{-v^2}{\rm d}v=(\sqrt{\pi/3})(\sqrt{\pi})=\frac{\pi}{\sqrt3}$$

For those who enjoy neat tricks, a probably not so helpful tour-de-force solution, stimulated by @mathbeing:
Write the exponent as a quadratic form
$$I=\iint_{\mathbb{R}^2} e^{-{\bf r}g{\bf r}}d^2 {\bf r}$$
where $g=\begin{bmatrix}2 & 1\\1 & 2\end{bmatrix}$ is the metric tensor. Write $g=L^T L$, and use ${\bf p}=L{\bf r}$ as substitution. Any valid $L$ will do, we don't need to compute it. We know $\det J =(\det L)^{-1} = \frac{1}{\sqrt{\det g}}=\frac{1}{\sqrt3}$.
$$I=\frac{1}{\sqrt3}\iint_{\mathbb{R}^2} e^{-|p|^2}d^2 {\bf p}\underbrace{=}_{\text{polar}}\frac{1}{\sqrt3}2\pi\int_0^\infty p e^{-p^2}dp=\frac{\pi}{\sqrt3}$$
To completely generalize, in any number of dimensions, for any possible nondegenerate positive definite quadratic form in the exponent, we have
$$\iiint_{\mathbb{R}^N}e^{-{\bf r}g{\bf r}}d^N{\bf r}=\pi^{N/2} (\det g)^{-1/2}$$
