Double Integral related to Gaussian Integral.

Question

We know that $\int_{-\infty}^{\infty} e^{-x^2}dx=\sqrt{\pi}.$
Using this , how can you evaluate $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2+y^2+xy)}dxdy= ?$
Are there any standard tricks for integrals which are related to the gaussian integral ?

Answer 1

The matrix $A:=\pmatrix{1&\frac12\\\frac12&1}$ is positive definite with square root $$A^{\frac12} = \frac14 \pmatrix{\sqrt 2 + \sqrt 6 & -\sqrt2 + \sqrt 6\\ -\sqrt 2 + \sqrt 6 & \sqrt 2 + \sqrt 6}$$ And your integral with $x := \pmatrix{x_1\\x_2}$ is $$\int_{\mathbb R^2} e^{-x^T A x} \ \mathrm dx$$ Now substitute $u = A^{\frac12} x$ and use a known integral.
Remark: $\det A^{\frac12} = \sqrt{\det A} = \frac{\sqrt 3}2$.

Answer 2

Hint

What about writing first $$x^2+x\,y+y^2=\big(x+\frac y2\big)^2+\frac 34 y^ 2$$ So, $$\int e^{-(x^2+xy+y^2)}\,dx=e^{-\frac 34 y^ 2}\int e^{-(x+\frac y2)^2}\,dx=e^{-\frac 34 y^ 2}\int e^{-z^2}\,dz$$ Use the bounds and the known $\int_{-\infty}^{\infty} e^{-z^2}\,dz=\sqrt{\pi}$ to continue.

I am sure that you can take from here.