What is the relationship among $a, b$ and $c$ for $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\left(ax^2+2bxy+cy^2\right)}dx dy=1$

Question

What relationship must hold between the constants $a, b$ and $c$ to make: $$ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{\rm e}^{-\left(\, ax^{2}\ +\ 2bxy\ +\ cy^2\,\right)}\phantom{A\,}{\rm d}x\,{\rm d}y = 1 $$

I am absolutely clueless on how to proceed with this question. I found a solution to this question on this website but the initial steps where it uses the transformation with the constraints on $\alpha, \beta, \gamma$ and $\delta$ is not clear to me. Like what was the intuition behind this transformation $?$. I understand that the end result is somewhat similar to the initial assumptions, but how am I supposed to think of this particular transformation in an exam?

I would appreciate any alternate answers/techniques to this question. Explanation(s) to the external linked solution are also welcome.

Edit #1: As mentioned by fellow users in the comments, the link is hidden behind a paywall. Here's the crux of what was given as the solution:

They used the transformation $$s=\alpha x+\beta y$$ $$t=\gamma x+\delta y$$ where $\left(\alpha\delta-\beta\gamma\right)^2=ac-b^2$.

Then they solved for $x$ and $y$ and proceeded with the Jacobian of the transformation and hence the given integral to obtain the required condition.

Answer 1

Recognise that the integral is proportional to the pdf for a bivariate normal distribution centred on the origin with inverse-of-covariance matrix $$\mathbf\Sigma^{-1}=2\begin{bmatrix}a&b\\b&c\end{bmatrix},\det\mathbf\Sigma=\frac1{4(ac-b^2)}$$ We require the normalisation $\frac1{2\pi}\frac1{\sqrt{\det\mathbf\Sigma}}$ to be $1$ to match the given integral, which becomes $$\det\mathbf\Sigma=\frac1{4\pi^2}=\frac1{4(ac-b^2)}\implies\pi^2=ac-b^2$$ Rewriting this as $\pi^2+b^2=ac$ we see that $a$ and $c$ are both positive or both negative. It cannot be the latter since the integral then diverges, so the conditions for the integral to be $1$ are $\pi^2=ac-b^2$ and $a,c>0$.

Answer 2

Integrate in polar coordinates as follows \begin{align} &\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\left(ax^2+2bxy+cy^2\right)}dx dy\\ =& \int_0^{2\pi}\int_0^\infty e^{-r^2\left(a\cos^2t+2b \cos t\sin t+c\sin^2t\right)}rdrdt \\ =& \ \frac12 \int_0^{2\pi} \frac1{a\cos^2t+2b \cos t\sin t+c\sin^2t}dt\\ =& \ \frac12 \int_0^{2\pi} \frac{\sec^2 t}{a+2b \tan t+c\tan^2t}dt\\ =&\ \frac1{2c} \int_0^{2\pi} \frac{d(\tan t)}{(\tan t+\frac bc)^2 + \frac{ac-b^2}{c^2}} = \frac1{c} \int_{-\pi/2}^{\pi/2} \frac{d(\tan t)}{\tan^2 t+ \frac{ac-b^2}{c^2}}\\ =&\ \frac1{\sqrt{ac-b^2}}\tan^{-1}\frac{c\tan t}{\sqrt{ac-b^2}}\bigg|_{-\pi/2}^{\pi/2}=\frac\pi{\sqrt{ac-b^2}}=1 \end{align} which leads to $ac-b^2=\pi^2$.

Answer 3

An alternative method is to complete the square for the x-variable as $(x+\frac{b}{a}y)^{2}$ and integrate over $x$, yielding a Gaussian-type integral: \begin{equation} \int_{-\infty}^{\infty}e^{-a((x+\frac{b}{a}y)^2-(\frac{b}{a}y)^2)}dx= \sqrt\frac{\pi}{a}e^{\frac{b^2}{a}y^2} \end{equation} Upon inserting the RHS into the y-variable integral, performing a second Gaussian integration, and noticing that convergence is ensured only if $c-\frac{b^2}{a}>0$, yields the result: \begin{equation} \frac{\pi}{\sqrt(ac-b^2)}= 1 \end{equation} Indeed, since $ac=\pi^{2}+b^2$, and since neither $a$ or $c$ can be negative for convergence reasons, both $a$ and $c$ must be simultaneously positive.