Evaluate $\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-5x^2-5y^2+8xy} dxdy$. Hint $\int_{-\infty}^{\infty}e^{-x^2}dx = \sqrt{\pi}$
The options are as follows:
- $\frac{\pi}{2}$
- $\frac{\pi}{3}$
- $\frac{\pi}{4}$
- $\frac{\pi}{5}$
- $\frac{\pi}{6}$
- $\frac{\pi}{8}$
My attempt:
$x = r\cos\theta$ and $y = r\sin\theta$
$x^2 + y^2 = r^2$
limits: $0<\theta<2\pi$ and $-\infty<r<\infty$
The integral expression becomes:
$$\int_0^{2\pi}\int_{-\infty}^\infty e^{-r^2(5-8\sin\theta\, \cos\theta)} drd\theta = \int_0^{2\pi}\frac {\sqrt{\pi}}{(5-8\sin\theta\, \cos\theta)}d\theta$$
This evaluates to
Let $u = 5-8\sin\theta\, \cos\theta = 5-4\sin(2\theta)$ hence $du = -8\cos(2\theta) d\theta$ and
$$\int_0^{2\pi}\frac {\sqrt{\pi}}{(5-8\sin\theta \, \cos\theta)}d\theta = \sqrt{\pi}\int_0^{2\pi}\frac {1}{u(-8\cos(2\theta))}du = -\frac{\sqrt{\pi}}{8\cos(2\theta)}.\ln(5-4\sin(2\theta)).$$
Applying the limits gives 0 as answer which is clearly wrong.
Best Answer
Setting $$ x=\frac{3u-v}{3\sqrt{2}},\quad y=\frac{3u+v}{3\sqrt{2}}, $$ the expression $-5(x^2+y^2)+8xy$ becomes \begin{eqnarray} -5(x^2+y^2)+8xy &=& -5\frac{9u^2-6uv+v^2+9u^2+6uv+v^2}{18}+8\frac{9u^2-v^2}{18}\\ &=&-5\frac{18u^2+2v^2}{18}+8\frac{9u^2-v^2}{18}\\ &=&-5\frac{9u^2+v^2}{9}+4\frac{9u^2-v^2}{9}\\ &=&\frac{(-45+36)u^2+(-5-4)v^2}{9}\\ &=&\frac{-9u^2-9v^2}{9}\\ &=&-(u^2+v^2) \end{eqnarray} The Jacobian matrix of the transformation $$ \Phi:(u,v) \mapsto (x,y)=\left(\frac{3u-v}{3\sqrt{2}},\frac{3u+v}{3\sqrt{2}}\right) $$ is $$ J_\Phi(u,v)=\begin{pmatrix} \frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u}&\frac{\partial y}{\partial v} \end{pmatrix} =\begin{pmatrix} \frac{1}{\sqrt{2}}&-\frac{1}{3\sqrt{2}}\\ \frac{1}{\sqrt{2}}&\frac{1}{3\sqrt{2}} \end{pmatrix} $$ It follows that $$ \det(J_\Phi(u,v))=\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{3\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{3\sqrt{2}}\right)=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}. $$ Hence \begin{eqnarray} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-5x^2-5y^2+8xy}dxdy&=&\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{3}e^{-u^2-v^2}dudv\\ &=&\frac{1}{3}\left(\int_{-\infty}^{\infty}e^{-u^2}du\right)\left(\int_{-\infty}^{\infty}e^{-v^2}du\right)\\ &=&\frac{1}{3}\sqrt{\pi}\sqrt{\pi}\\ &=&\frac{\pi}{3} \end{eqnarray}