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?

  • $\begingroup$ Where are the limits of integration on $y$? $\endgroup$ – Shubham Johri Jan 12 at 18:17
  • 1
    $\begingroup$ How can you separate this integral into those two integrals? The integrand of the first integral is the product of the integrands of the latter two, not the sum. $\endgroup$ – Frpzzd Jan 12 at 18:19
  • $\begingroup$ Are you sure that what you want isn't $\displaystyle\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(3x^2+2\sqrt2xy+3y^2)}\,\mathrm dx\,\mathrm dy$? $\endgroup$ – José Carlos Santos Jan 12 at 18:23
  • $\begingroup$ Sorry I have corrected my problem. $\endgroup$ – Dbchatto67 Jan 12 at 18:23

$$3x^2+2\sqrt{2} xy + 3y^2 =\begin{bmatrix}x & y \end{bmatrix} \begin{bmatrix} 3 & \sqrt{2} \\ \sqrt{2} & 3 \end{bmatrix} \begin{bmatrix}x \\ y \end{bmatrix}$$ 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 $$\frac{1}{\sqrt{(2 \pi)^2 \det (\Omega^{-1})}} \int_{-\infty}^\infty \int_{-\infty}^\infty \exp(-v^\top \Omega v / 2) \, dx \, dy = 1.$$

  • $\begingroup$ fascinating . . . definitely learned something new $\endgroup$ – Chase Ryan Taylor Jan 12 at 18:37
  • $\begingroup$ 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? $\endgroup$ – Chase Ryan Taylor Jan 13 at 6:46
  • $\begingroup$ @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. $\endgroup$ – angryavian Jan 13 at 6:48
  • $\begingroup$ If one writes $v=xi+yj$ then $dv=dx\,i + dy\,j$, no? Which is different? $\endgroup$ – Chase Ryan Taylor Jan 13 at 6:49

Hint: Doing the change of variables $x=X+Y$ and $y=X-Y$, your integral becomes$$2\int_{-\infty}^\infty\int_{-\infty}^\infty \exp\left[-\left(2\sqrt2+6\right)X^2-\left(-2\sqrt2+6\right)Y^2\right]\,\mathrm dX\,\mathrm dY.$$


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