Crazy Gaussian Integral

Change to polar co-ordiantes :

$x=rcos(\theta)$

$y=rsin(\theta)$

And our area element $dxdy$ becomes $rd\theta dr$

Also we are integrating it for the whole space :

Integral becomes :

$\large \int _{ 0 }^{ 2\pi }{ \int _{ 0 }^{ \infty }{ r{ e }^{ -{ r }^{ 2 }(a{ cos }^{ 2 }(\theta )+2bcos(\theta )sin(\theta )+c{ sin }^{ 2 }(\theta )) }drd\theta } }$

With respect to dr this can be easily solved by substituting ${r}^{2}=t$

Integrating first integral we get :

$\int _{ 0 }^{ 2\pi }{ \frac { 1 }{ 2(a{ cos }^{ 2 }(\theta )+2bcos(\theta )sin(\theta )+c{ sin }^{ 2 }(\theta )) } d\theta }$

Which can be written as :

$I=\int _{ 0 }^{ \pi }{ \frac { 1 }{ (a{ cos }^{ 2 }(\theta )+2bcos(\theta )sin(\theta )+c{ sin }^{ 2 }(\theta )) } d\theta }$

Breaking it in parts we get :

$I=\int _{ 0 }^{ \pi /2 }{ \frac { 1 }{ (a{ cos }^{ 2 }(\theta )+2bcos(\theta )sin(\theta )+c{ sin }^{ 2 }(\theta )) } d\theta } +\int _{ \pi /2 }^{ \pi }{ \frac { 1 }{ (a{ cos }^{ 2 }(\theta )+2bcos(\theta )sin(\theta )+c{ sin }^{ 2 }(\theta )) } d\theta }$

Solving both integrals (as they are pretty standard and can be solved by first dividing numerator and denominator by ${cos}^{2}(\theta)$ and substituting $tan(\theta)=t$ ) we get :

$\frac { \pi }{ \sqrt { ac-{ b }^{ 2 } } } =I=1$

$\Rightarrow ac-{b}^{2}={\pi}^{2}$

Just use gaussian distribution: $\int\limits_{-\infty }^{\infty }{\frac{1}{\sqrt{2\pi {{\sigma }^{2}}}}{{e}^{-\frac{{{(x-\mu )}^{2}}}{2{{\sigma }^{2}}}}}dx}=1$

Knowing that: $\int\limits_{-\infty }^{\infty }{{{e}^{-\alpha {{(x-\mu )}^{2}}}}dx}=\sqrt{\frac{\pi }{\alpha }}$ The Integral may be rewriteen as: $\int\limits_{-\infty }^{\infty }{\int\limits_{-\infty }^{\infty }{{{e}^{-(a{{x}^{2}}+2byx+c{{y}^{2}})}}dxdy}}=\int\limits_{-\infty }^{\infty }{\int\limits_{-\infty }^{\infty }{{{e}^{-a{{(x+\frac{b}{a}y)}^{2}}}}{{e}^{-(c-\frac{{{b}^{2}}}{a}){{y}^{2}}}}dxdy}}=$ $\sqrt{\frac{\pi }{a}}\int\limits_{-\infty }^{\infty }{{{e}^{-(c-\frac{{{b}^{2}}}{a}){{y}^{2}}}}dy}=\sqrt{\frac{\pi }{a}}\sqrt{\frac{\pi }{c-\frac{{{b}^{2}}}{a}}}=\sqrt{\frac{{{\pi }^{2}}}{ac-{{b}^{2}}}}=1$

Therefore: $ac-{{b}^{2}}={{\pi }^{2}}$