Integral of a two variable function $e^{-(5x^2-6xy+5y^2)}$ How to compute $$\int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(5x^2-6xy+5y^2)} \, dx \, dy \;?$$
I have seen the integral first time. I can't find any hint how to even proceed? Kindly HELP. Thank You!
 A: Assume that $Ax^2+Bxy+Cy^2$ is a positive definite quadratic form, i.e. $A>0$ and $B^2<4AC$. By diagonalizing such quadratic form, we get that
$$\iint_{\mathbb{R}^2}\exp\left[-(Ax^2+Bxy+Cy^2)\right]\,dx\,dy = \iint_{\mathbb{R}^2}\exp(-\lambda_1 x^2-\lambda_2 y^2)\,dx\,dy $$
with $\lambda_1,\lambda_2$ being the eigenvalues of $M=\begin{pmatrix} A & B/2 \\ B/2 & C \end{pmatrix}$. By Fubini's theorem the last integral equals
$$ \frac{\pi}{\sqrt{\lambda_1 \lambda_2}} = \frac{\pi}{\sqrt{\det M}}=\color{red}{\frac{2\pi}{\sqrt{4AC-B^2}}}. $$
A: Try to use the fact that the integral of the density of the normal distribution is 1:
$$\frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac{(x-\mu)^2}{2\sigma^2}} \, dx=1.$$
Here $-(5x^2-6xy+5y^2)=-5(x-\frac{3}{5}y)^2-\frac{16}{5}y^2$, so our integral is
$$\int_{-\infty}^\infty e^{-\frac{16}{5}y^2}\int_{-\infty}^\infty e^{-5(x-\frac{3}{5}y)^2} \, dx\,dy=\int_{-\infty}^\infty e^{-\frac{16}{5}y^2}\frac{\sqrt{\pi}}{\sqrt{5}}\frac{1}{\frac{1}{\sqrt{10}}\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac{(x-\frac{3}{5}y)^2}{2\cdot(\frac{1}{\sqrt{10}})^2}}\,dx\,dy.$$
Here $$\frac{1}{\frac{1}{\sqrt{10}}\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac{(x-\frac{3}{5}y)^2}{2\cdot(\frac{1}{\sqrt{10}})^2}}dx=1,$$ so our integral is
$$\frac{\sqrt{\pi}}{\sqrt{5}}\int_{-\infty}^\infty e^{-\frac{16}{5}y^2}dy=\frac{\sqrt{\pi}}{\sqrt{5}}\frac{\sqrt{2\pi}}{\frac{\sqrt{32}}{\sqrt{5}}}\frac{1}{\frac{\sqrt{5}}{\sqrt{32}}\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac{y^2}{2\cdot(\frac{\sqrt{5}}{\sqrt{32}})^2}}dy.$$
Here, again, $$\frac{1}{\frac{\sqrt{5}}{\sqrt{32}}\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac{y^2}{2\cdot(\frac{\sqrt{5}}{\sqrt{32}})^2}}dy=1,$$ so our integral is
$$\frac{\sqrt{\pi}}{\sqrt{5}}\frac{\sqrt{2\pi}}{\frac{\sqrt{32}}{\sqrt{5}}}=\frac{\pi}{4}.$$
A: In $5x^2 - 6xy + 5y^2$, the coefficients of $x^2$ and $y^2$ are the same, which to me suggests that rotating the coordinate system to eliminate the $xy$ term should be done in a way that treats $x$ and $y$ equally. That would mean a $45^\circ$ rotation, thus:
\begin{align}
x & = u+v  \\
y & = u-v
\end{align}
so that
\begin{align}
5x^2 -6xy+5y^2 & = 5(u+v)^2 -6(u+v)(u-v) + 5(u-v)^2 \\[10pt]
& = 5(u^2+2uv+v^2) - 6(u^2-v^2) + 5(u^2-2uv+v^2) \\[10pt]
& = 4u^2 + 16v^2. 
\end{align}
And
$$
dx\,dy = \left| \det\left[ \begin{array}{cc} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\[6pt] \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v}  \end{array} \right] \right| \, du\, dv = 2\,du\,dv
$$
