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 ?

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    $\begingroup$ $x^2 + y^2 + xy = \frac12(x+y)^2 + \frac12(x^2 + y^2)$. $\endgroup$ – AlexR Feb 19 '15 at 11:20
  • $\begingroup$ @AlexR.That's nice. So,we have $e^-{({1\over2}(x+y)^2+{1\over2}(x^2+y^2))}$. I'm not able to see the next step . I tried to keep $z=x+y$Can you please give a hint ? $\endgroup$ – Srinivas K Feb 19 '15 at 11:45
  • $\begingroup$ There should be a more elegant solution, but I provided a nuke-a-fly solution. $\endgroup$ – AlexR Feb 19 '15 at 11:55
  • $\begingroup$ @AlexRwikipedia says that if $A$ is a positive definite symmetric covariance matrix , the value of integral must be $\sqrt{(2\pi)^n/det(A)}$. Then the answer would be $4\pi/\sqrt{3}$Is this what we need to use ? $\endgroup$ – Srinivas K Feb 19 '15 at 12:03
  • $\begingroup$ Yes, that result is obtained with the general approach I used in my answer via $2\pi \det(A^{-\frac12}) = \frac{2\pi}{\det A^{\frac12}} = \frac{2\pi}{\sqrt{\det A}}$. $\endgroup$ – AlexR Feb 19 '15 at 12:12

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: $\det A^{\frac12} = \sqrt{\det A} = \frac{\sqrt 3}2$.

  • $\begingroup$ Can you please give some reference on this topic? Is this related to matrix calculus? $\endgroup$ – Yuriy S Feb 26 '16 at 12:16


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.


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