<s>[INST] 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!
 [/INST] 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
$$\frac{\sqrt{\pi}}{\sqrt{5}}\frac{\sqrt{2\pi}}{\frac{\sqrt{32}}{\sqrt{5}}}=\frac{\pi}{4}.$$
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