# How do I evaluate $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(3x^2+2 \sqrt 2 xy+3y^2)} \mathrm dx\,\mathrm dy$?

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?

• Where are the limits of integration on $y$? – Shubham Johri Jan 12 at 18:17
• 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. – Frpzzd Jan 12 at 18:19
• Are you sure that what you want isn't $\displaystyle\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(3x^2+2\sqrt2xy+3y^2)}\,\mathrm dx\,\mathrm dy$? – José Carlos Santos Jan 12 at 18:23
• Sorry I have corrected my problem. – Dbchatto67 Jan 12 at 18:23
• – StubbornAtom Jan 12 at 20:04

$$3x^2+2\sqrt{2} xy + 3y^2 =\begin{bmatrix}x & y \end{bmatrix} \begin{bmatrix} 3 & \sqrt{2} \\ \sqrt{2} & 3 \end{bmatrix} \begin{bmatrix}x \\ y \end{bmatrix}$$ 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 $$\frac{1}{\sqrt{(2 \pi)^2 \det (\Omega^{-1})}} \int_{-\infty}^\infty \int_{-\infty}^\infty \exp(-v^\top \Omega v / 2) \, dx \, dy = 1.$$
• 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? – Chase Ryan Taylor Jan 13 at 6:46
• @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 at 6:48
• If one writes $v=xi+yj$ then $dv=dx\,i + dy\,j$, no? Which is different? – Chase Ryan Taylor Jan 13 at 6:49
Hint: Doing the change of variables $$x=X+Y$$ and $$y=X-Y$$, your integral becomes$$2\int_{-\infty}^\infty\int_{-\infty}^\infty \exp\left[-\left(2\sqrt2+6\right)X^2-\left(-2\sqrt2+6\right)Y^2\right]\,\mathrm dX\,\mathrm dY.$$