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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?

Anastasiya-Romanova 秀
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Debashish
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    This is odd, both strategies you mention actually work. What stopped you in each case? – Did Jul 25 '14 at 09:00
  • In the first case I am not being able to integrate $\int_{-\infty}^{\infty}e^{-\frac{1}{2}(x^2-xy+y^2)}dy$ – Debashish Jul 25 '14 at 09:06
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    Even after noting that $y^2-xy+x^2=(y-\frac12x)^2+\frac34x^2$ and using the change of variable $y=z+\frac12x$ to integrate on $y$? – Did Jul 25 '14 at 09:08
  • oh yes... thats good .. So I think I can then reduce it to the form $e^{-z^2}$.. Ok lets try. – Debashish Jul 25 '14 at 09:10
  • Let's, indeed... :-) – Did Jul 25 '14 at 09:20
  • https://math.stackexchange.com/questions/384732/find-the-value-of-int-infty-infty-int-infty-inftye-x2xyy2?noredirect=1&lq=1. – StubbornAtom Jun 15 '19 at 20:32

2 Answers2

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

Matthias
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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 $A=\left(\begin{array}{cc} 1 & -1/2 \\ -1/2 & 1\end{array}\right)$.

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