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Evaluate $$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \exp\left(-3x^2-2 \sqrt 2 xy - 3y^2\right) \, \mathrm dx\,\mathrm dy$$

I first evaluate

$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \exp\left[-3\bigl(x^2+ y^2\bigr)\right] \,\mathrm dx\,\mathrm dy$$

using polar coordinates, which evaluates to $\pi/3$. But I find difficulty to evaluate the double integral $$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \exp\left(-2 \sqrt 2 xy\right) \, \mathrm dx\,\mathrm dy$$ Would anybody please help me finding it out?

StubbornAtom
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little o
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2 Answers2

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$$3x^2+2\sqrt{2} xy + 3y^2 =\begin{bmatrix}x & y \end{bmatrix} \begin{bmatrix} 3 & \sqrt{2} \\ \sqrt{2} & 3 \end{bmatrix} \begin{bmatrix}x \\ y \end{bmatrix}$$ so the integrand is $$\exp(- v^\top \Omega v/2)$$ where $v = \begin{bmatrix}x \\ y \end{bmatrix}$ and $\Omega = 2\begin{bmatrix} 3 & \sqrt{2} \\ \sqrt{2} & 3 \end{bmatrix}$.

By using the density of a $N(0, \Sigma)$ distribution we have $$\frac{1}{\sqrt{(2 \pi)^2 \det (\Omega^{-1})}} \int_{-\infty}^\infty \int_{-\infty}^\infty \exp(-v^\top \Omega v / 2) \, dx \, dy = 1.$$

angryavian
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  • fascinating . . . definitely learned something new – gen-ℤ ready to perish Jan 12 '19 at 18:37
  • I've actually been studying this post and all the new concepts it's exposed me to. If the original integrand is $\exp\bigl(-v^\top\Omega v/2\bigr)$, then what allows $dv=d\begin{bmatrix}x \ y \end{bmatrix}$ to stand for $dx,dy$, from the original differential? – gen-ℤ ready to perish Jan 13 '19 at 06:46
  • @ChaseRyanTaylor If you wish, you may write out $v$ as $(x,y)$ and write the integral as $\int_{-\infty}^\infty \int_{-\infty}^\infty \cdots ,dx ,dy$. I was just using some shorthand, but it is exactly the same as the original integral. – angryavian Jan 13 '19 at 06:48
  • If one writes $v=xi+yj$ then $dv=dx,i + dy,j$, no? Which is different? – gen-ℤ ready to perish Jan 13 '19 at 06:49
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Hint: Doing the change of variables $x=X+Y$ and $y=X-Y$, your integral becomes$$2\int_{-\infty}^\infty\int_{-\infty}^\infty \exp\left[-\left(2\sqrt2+6\right)X^2-\left(-2\sqrt2+6\right)Y^2\right]\,\mathrm dX\,\mathrm dY.$$

gen-ℤ ready to perish
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José Carlos Santos
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