Evaluating $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}{ {{1}\over{2\pi}} e^{{{-1}\over{2}}(4x^2-2xy+3y^2)} ~\mathrm dx \mathrm dy}$ How to calculate that integral? I have no idea what substitution to use. I think polar coordinates will be needed here.

$$\displaystyle \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}{ {{1}\over{2\pi}} e^{{{-1}\over{2}}(4x^2-2xy+3y^2)} ~\mathrm dx \mathrm dy}$$

 A: You can even do it with $(x,y)$. Complete the square first
$$\int{ \frac{1}{2\pi}} e^{{{-1}\over{2}}(4x^2-2xy+3y^2)} dx=\frac{e^{-\frac{11 y^2}{8}} \text{erf}\left(\frac{4 x-y}{2 \sqrt{2}}\right)}{4
   \sqrt{2 \pi }}$$
$$\int_{-\infty}^\infty{ \frac{1}{2\pi}} e^{{{-1}\over{2}}(4x^2-2xy+3y^2)} dx=\frac{e^{-\frac{11 y^2}{8}}}{2 \sqrt{2 \pi }}$$
$$\int \frac{e^{-\frac{11 y^2}{8}}}{2 \sqrt{2 \pi }}\,dy=\frac{\text{erf}\left(\frac{1}{2} \sqrt{\frac{11}{2}} y\right)}{2 \sqrt{11}}$$
$$\int_{-\infty}^\infty \frac{e^{-\frac{11 y^2}{8}}}{2 \sqrt{2 \pi }}\,dy=\frac{1}{\sqrt{11}}$$
In fact, for the most general case
$$I=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\frac 1{2\pi} e^{-(ax^2+bxy+by^2)}\,dx\,dy=\frac{1}{\sqrt{4 a c-b^2}}$$ if
$$\Re(a)>0\land \Re\left(\frac{b^2}{a}\right)<4 \Re(c)$$
A: Hint
First prove that your integral converges. Then $$\frac{1}{2\pi}\iint_{\mathbb R^2}e^{-\frac{1}{2}(4x^2-2xy+3y^2)}\,\mathrm d x\,\mathrm d y=\lim_{n\to \infty }\iint_{\mathcal B(0,n)}e^{-\frac{1}{2}(4x^2-2xy+3y^2)}\,\mathrm d x\,\mathrm d y,$$
where $\mathcal B(0,n)=\{(x,y)\mid x^2+y^2\leq n\}$. Using polar coordinates allow you to conclude.
