limit of integration after change of variables Evaluate $$\int_{0}^{\infty}\int_{0}^{\infty}e^{-(5x^2-6xy+5y^2)}dxdy$$
after applying the change of variables as $$x=u+v~~,y=u-v$$
i got the integral as $$\int\int e^{-4u^2-16v^2}{2}dudv$$
But how do i find the limits of integration?
After seeing the cooments i got the integral set up as $$\int_{-\infty}^{\infty}\int_{0}^{\infty} e^{-4u^2-16v^2}{2}dudv=4\int_{0}^{\infty}\int_{0}^{\infty} e^{-4u^2-16v^2}dudv=4\int_{0}^{\infty}e^{-4u^2}du\int_{0}^{\infty}e^{-16v^2}dv=\dfrac{\pi}{8}$$, 
But when i solve this integral from wolfram it gives $\dfrac{1}{16}\left(\pi+2\tan^{-1}\dfrac{3}{4}\right)$,
i can't find the error in my calculation, can somebody help
 A: hint
$$u=\frac {x+y}{2} $$
$$v=\frac {x-y}{2} $$
$$0\le x<+\infty \land 0\le  y<+\infty$$
$$\implies 0\le  u<+\infty  \land -\infty <v <+\infty$$
A: The boundary $x=0$, $y\ge 0$ in the $x-y$ plane maps to the boundary $u=-v$, with $u\in [0,\infty)$ and $v\in (-\infty,0]$ in the $u-v$ plane.
The boundary $y=0$, $x\ge 0$ in the $x-y$ plane maps to the boundary $u=v$, with $u\in [0,\infty)$ and $v\in [0,\infty)$ in the $u-v$ plane.
Therefore, the region in the first quadrant in the $x-y$ plane maps to the region bounded by $u=-v$ and $u=v$, $u\ge 0$ in the $u-v$ plane.
The integral becomes
$$\int_0^\infty \int_0^\infty f(x,y)\,dx\,dy=\int_0^\infty\int_{-u}^u f(u+v,u-v)\,J(u,v)\,dv \,du$$
where $J(u,v)=\left|\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}\right|=2$ is the Jacobian of transformation.
A: 
Note that lines of constant $u$ run bottom right to top left & are indicated in red,blue & green.
Lines of constant $v$ run bottom left to top right & are indicated in yellow, purple & black.
For the red line ($u=1$) $v$ varies from $-1$ to $1$ and more generally for $u$, $v$ will vary from $-u$ to $u$ so
\begin{eqnarray*}
\int_0^{\infty} dx \int_0^{\infty} dy = \int_0^{\infty} du \int_{-u}^{u} dv 
\end{eqnarray*}
And you will also need to Multiply by the Jacobian.
A: We have the integral
$$\int\int_Re^{-4u^2-16v^2}2du\,dv$$
The $x$-axis becomes $y=u-v=0$. Since $u=(x+y)/2\ge0$ this boundary becomes the half-line $v=u$ for $u\ge0$.
The $y$-axis becomes $x=u+v=0$, so this boundary becomes the half-line $v=-u$ for $u\ge0$.
Transform to polar coordinates $u=r\cos\theta$, $v=r\sin\theta$ and $u+v=0$ becomes $\theta=-\pi/4$, $u-v=0$ becomes $\theta=\pi/4$ so the integral becomes
$$\begin{align}\int\int_Re^{-4u^2-16v^2}2du\,dv&=2\int_{-\pi/4}^{\pi/4}\int_0^{\infty}e^{-r^2\left(4\cos^2\theta+16\sin^2\theta\right)}r\,dr\,d\theta\\
&=\int_{-\pi/4}^{\pi/4}\left[\frac{-e^{-r^2\left(4\cos^2\theta+16\sin^2\theta\right)}}{4\cos^2\theta+16\sin^2\theta}\right]_0^{\infty}d\theta\\
&=\int_{-\pi/4}^{\pi/4}\frac{d\theta}{4\cos^2\theta+16\sin^2\theta}=\int_{-1}^1\frac{du}{4+16u^2}\\
&=\int_{-\tan^{-1}2}^{\tan^{-1}2}\frac18d\phi=\frac14{\tan^{-1}2}\end{align}$$
Where we have made the substitutions $u=\tan\theta$ and $2u=\tan\phi$. Wolfram Alpha probably got its answer by directly transforming to polar coordinates, skipping the $(u,v)$ transformation. The two answer are the same even though they look different. Since $2\tan^{-1}2$ is a second-quadrant angle,
$$\begin{align}\frac14\tan^{-1}2&=\frac18\left(2\tan^{-1}2\right)=\frac18\left(\tan^{-1}\left(\tan\left(2\tan^{-1}2\right)\right)+\pi\right)\\
&=\frac18\left(\tan^{-1}\left(\frac{2(2)}{1-(2)^2}\right)+\pi\right)=\frac18\left(\pi-\tan^{-1}\frac43\right)=\frac18\left(\pi-\cot^{-1}\frac34\right)\\
&=\frac18\left(\pi-\left(\frac{\pi}2-\tan^{-1}\frac34\right)\right)=\frac18\left(\frac{\pi}2+\tan^{-1}\frac34\right)\end{align}$$
