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 $x$ &amp; $y$. 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?
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
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}