Calculate $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} dx dy \exp(-5x^2+8xy-5y^2)$

Question

I need to evaluate the following integral $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} dx dy \exp(-5x^2+8xy-5y^2)$$

I know the normal double integrals but this one seem different, moreover, I'm familiar with $$\int_{-\infty}^{\infty} \exp(-x^2)dx = \sqrt\pi$$ but I cannot figure out the polar substitution for $-5x^2+8xy-5y^2$ expression. I tried substituting $x^2+y^2$ at $-5(x^2+y^2)$ with $r^2$ and $8xy$ with $r^2\sin\theta\cos\theta$. And then I got even more lost.

Can someone please help me figure this question out and is there any general solution for such problems? And also can someone provide me a link to similar problems and their solutions in order for me to understand better. Thank you!

Answer 1

If you make the substitution $x=\frac{a+b}{\sqrt2}$ and $y=\frac{a-b}{\sqrt2}$, then your integral becomes\begin{align}\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-a^2-9b^2}\,\mathrm db\,\mathrm da&=\left(\int_{-\infty}^\infty e^{-a^2}\mathrm da\right)\left(\int_{-\infty}^\infty e^{-9b^2}\mathrm db\right)\\&=\sqrt\pi\frac{\sqrt\pi}3\\&=\frac\pi3.\end{align}

Answer 2

To begin, we complete the square to notice that

$$-5x^{2}+8xy-5y^{2}=-5\left(x-\frac{4}{5}y\right)^{2}-\frac{9}{5}y^{2}$$

so we get that

\begin{align*} &\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp(-5x^{2}+8xy-5y^{2})dxdy\\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp\left(-5\left(x-\frac{4}{5}y\right)^{2}-\frac{9}{5}y^{2}\right)dxdy.\tag{1} \end{align*}

For any value of $y$ it holds that

$$\int_{-\infty}^{\infty}\exp\left(-5\left(x-\frac{4}{5}y\right)^{2}\right)dx=\int_{-\infty}^{\infty}\exp\left(-5x^{2}\right)dx$$

because of the substiuttion $u=x-\frac{4}{5}y$, so substituting back into (1) we get that

\begin{align*} &\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp(-5x^{2}+8xy-5y^{2})dxdy\\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp\left(-5x^{2}-\frac{9}{5}y^{2}\right)dxdy\\ &=\left(\int_{-\infty}^{\infty}\exp\left(-5x^{2}\right)dx\right)\left(\int_{-\infty}^{\infty}\exp\left(-\frac{9}{5}y^{2}\right)dy\right) \end{align*}

Now making the subs $u=\sqrt{5}x$ and $u=\frac{3}{\sqrt{5}}y$ we get that

\begin{align*} &\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp(-5x^{2}+8xy-5y^{2})dxdy\\ &=\left(\frac{1}{\sqrt{5}}\int_{-\infty}^{\infty}\exp\left(-x^{2}\right)dx\right)\left(\frac{\sqrt{5}}{3}\int_{-\infty}^{\infty}\exp\left(-y^{2}\right)dy\right)\\ &=\left(\frac{\sqrt{\pi}}{\sqrt{5}}\right)\left(\frac{\sqrt{5\pi}}{3}\right)\\ &=\frac{\pi}{3} \end{align*}

So we are done