POTW Gaussian Integrals in Two Dimensions

[Euge]

Let $a, b$, and $c$ be real numbers such that $a$ and $c$ are positive and $ac > b^2$. Evaluate the double integral $$\int_{-\infty}^\infty \int_{-\infty}^\infty e^{-ax^2 - 2bxy - cy^2}\, dx\, dy$$

 

[pasmith]

Diagonalizing \begin{pmatrix} a & b \\ b & c \end{pmatrix} seems like a good start.

 

[anuttarasammyak]

Spoiler

The matrix in post 2 is symmetric so it is diagonalized by multiplications of orthonormal matrix and its inversed so transversed matrix from the both ends. The orthonormal matrix change bases from x,y to another orthonormal bases, say u,v with $dxdy=dudv$.
<br /> ax^2+2bxy+cy^2=\begin{pmatrix}<br /> x &amp; y \\<br /> \end{pmatrix}<br /> \begin{pmatrix}<br /> a &amp; b \\<br /> b &amp; c \\<br /> \end{pmatrix}<br /> \begin{pmatrix}<br /> x \\<br /> y \\<br /> \end{pmatrix}<br /> =<br /> \begin{pmatrix}<br /> u &amp; v \\<br /> \end{pmatrix}<br /> \begin{pmatrix}<br /> \lambda_1&amp; 0 \\<br /> 0 &amp; \lambda_2 \\<br /> \end{pmatrix}<br /> \begin{pmatrix}<br /> u \\<br /> v \\<br /> \end{pmatrix}<br /> =\lambda_1 u^2+ \lambda_2 v^2<br />

In this new bases the double integral is carried out independently so the integral is
\frac{\pi}{\sqrt{\lambda_1\lambda_2}}
where eigenvalues $\lambda_1,\lambda_2$ are solutions of quadratic secular equation
(a-\lambda)(c-\lambda)-b^2=0
\lambda^2 -(a+c)\lambda+ac-b^2=0
Thus the integral is
\frac{\pi}{\sqrt{ac-b^2}}

 

[julian]

Spoiler

\begin{align*}
\int_{-\infty}^\infty \int_{-\infty}^\infty e^{-ax^2 - 2bxy - cy^2} dxdy & = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-a \left( x + \frac{b}{a} y \right)^2 + \frac{b^2}{a} y^2 - cy^2} dxdy
\nonumber \\
& = \int_{-\infty}^\infty \left( \int_{-\infty}^\infty e^{-a \left( x + \frac{b}{a} y \right)^2} dx \right) e^{- \frac{1}{a} (ac - b^2) y^2} dy
\nonumber \\
& = \sqrt{\frac{\pi}{a}} \int_{-\infty}^\infty e^{- \frac{1}{a} (ac - b^2) y^2} dy \qquad (\text{note: } ac-b^2 > 0)
\nonumber \\
& = \sqrt{\frac{\pi}{a}} \cdot \sqrt{\frac{a \pi}{ac-b^2}}
\nonumber \\
& = \frac{\pi}{\sqrt{ac-b^2}}
\end{align*}