limit of integration after change of variables

Question

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

Answer 1

hint

$$u=\frac {x+y}{2} $$ $$v=\frac {x-y}{2} $$

$$0\le x<+\infty \land 0\le y<+\infty$$ $$\implies 0\le u<+\infty \land -\infty <v <+\infty$$

Answer 2

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.

Answer 3

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.