Value of an unbounded definite integral Evaluate the integration :
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(2x^{2}+2xy+2y^{2})}dxdy$$
The function is even about $x$ & $y$. So we can write, $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(2x^{2}+2xy+2y^{2})}dxdy=2\int_{0}^{\infty}\int_{0}^{\infty}e^{-(2x^{2}+2xy+2y^{2})}dxdy$$.
But now , how I will proceed it?
 A: Take the integral 
$$
I=\int_R\int_R dxdy e^{-\frac{1}{2}\vec{x}^t M \vec{x}} 
$$
where $M$ is a positiv definite and symmetric $2\times2$ Matrix. In this case we can find an orthogonal transformation $S$ that fullfills $S^{t}MS=D$ where D is a diagonal matrix. Using $S^tS=1$ we can rewrite the exponent as $\underbrace{\vec{x}^tS}_{\vec{\zeta}^t}\underbrace{S^tMS}_{D}\underbrace{S^t\vec{x}}_{\vec{\zeta}}$, where the entries of $D$ are the two eigenvalues of $M$. The measure transforms as $dxdy \rightarrow \underbrace{|\det(S)|}_{1}d\zeta_1d\zeta_2$ so the whole integral reads:
$$
I=\int_R\int_R d\zeta_1d\zeta_2 e^{-\frac{1}{2}\vec{\zeta}^t D \vec{\zeta}}= \int_R\int_R d\zeta_1d\zeta_2 e^{-\frac{1}{2}\zeta_1 \lambda_1 \zeta_1}e^{-\frac{1}{2}\zeta_2 \lambda_2 \zeta_2}
$$
Which is just the product of two independent Gaussian integrals. It follows that I is just given by
$$
I=\sqrt{\frac{2\pi}{\lambda_1}}\sqrt{\frac{2\pi}{\lambda_2}}=\frac{2\pi}{\sqrt{\det(M)}} 
$$
Which is in your case $(\lambda_1=2,\lambda_1=6)$ 
$$
\pi \frac{1}{\sqrt{3}}
$$
This method is generalizable to non symmetric positive definite $n\times n$ matirx by using the decomposition $2M=(M+M^t)+(M-M^t)$. 
Edit: Correct Eigenvalues used
A: If you substitute $x=u+v$ and $y=u-v$ you get $$x^2+xy+y^2 = (u+v)^2+(u+v)(u-v)+(u-v)^2 = 3u^2+v^2.$$ In order to do the change of variables you will need to compute the Jacobian for this change of variables. In this case $u=\frac{x+y}{2}$ and $v=\frac{x-y}{2}$ and the Jacobian matrix is given by $$\left[ \begin{array}{cc}1/2 & 1/2\\1/2 &-1/2\end{array}\right]$$ and its determinant is $-1/4-1/4=-1/2$.
Now you can rewrite the integral as $$\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-2(3u^2)-2v^2} |-1/2| dudv = 1/2 \int_{-\infty}^\infty e^{-2(3u^2)}du \int_{-\infty}^\infty e^{-2v^2}dv.$$ Now if you know the integral formula for a Gaussian, you can plug that in to these two integrals. Recall that $\int_{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi}$.
A: Hint: $$2x^2+2xy+2y^2=2\bigl(x+\tfrac{1}{2}y\bigr)^2+\frac{3}{2}y^2.$$ Now use change of variable $u=\sqrt{2}(x+y/2)$ and $v=\sqrt{3/2}y$
A: Assuming $x$ and $y$ are real, Wolfram gives the indefinite integral as
\begin{equation}
\frac{1}{2} e^{-2y^{2}} \sqrt{e^{y^{2}}} \sqrt{\frac{\pi}{2}} erf\left(\frac{2x+y}{\sqrt{2}}\right)
\end{equation}
