Evaluate $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\frac{1}{2}(x^2-xy+y^2)}dx\, dy$

I need to evaluate the following integral:

$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\frac{1}{2}(x^2-xy+y^2)}dx\, dy$$

I thought of evaluating the iterated integral $\displaystyle\int_{-\infty}^{\infty}dx\int_{-\infty}^{\infty}e^{-\frac{1}{2}(x^2-xy+y^2)}dy$, but because of the presence of $x^2$ and $y^2$ terms, I am not being able to do that. I tried substituting $x=r\cos \theta$ and $y=r\sin \theta$ but in that case I have some confusion regarding the limits of $r$ and $\theta$. Can I get some help?

Solution 1:

Recall: $$x^2-xy+y^2=(x-\frac{1}{2}y)^2+\frac{3}{4}y^2$$ With that you get: $$\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1}{2}(x^2-xy+y^2)}dxdy=\\\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1}{2}\left((x-\frac{1}{2}y)^2+\frac{3}{4}y^2\right)}dxdy=\\ \int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1}{2}\left(x'^2+\frac{3}{4}y^2\right)}dx'dy=\\ \int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1}{2}x'^2}dx'e^{-\frac{3}{8}y^2}dy=\\ \int_{-\infty}^\infty\sqrt{2\pi}e^{-\frac{3}{8}y^2}dy=\\ \sqrt{2\pi}\sqrt{\frac{8}{3}\pi}=\frac{4}{\sqrt{3}}\pi $$

Solution 2:

The general gaussian integration formula is $$\int_{-\infty}^{\infty}\ldots\int_{-\infty}^{\infty} e^{-\frac12 x^T Ax}dx_1\ldots dx_n=\frac{(2\pi)^{\frac{n}{2}}}{\sqrt{\operatorname{det}A}}.$$ In your case, $n=2$ and $A=\left(\begin{array}{cc} 1 & -1/2 \\ -1/2 & 1\end{array}\right)$.