Evaluate $\iint (\frac{(x-y)}{x+y})^4$ over the triangular region bounded by $x+y=1$, $x$-axis ,$y$-axis 
Evaluate $\iint (\frac{(x-y)}{x+y})^4$ over the triangular region bounded by $x+y=1$, $x$-axis, $y$-axis.

My attempt: I tried using the change of variable concept:
Let $u=x-y$ and $v=x+y$ ,$|J|= \frac{1}{2}$
Then $x=\frac{u+v}{2}$ and $y=\frac{u-v}{2}$
Now we have to find the limit which is where I am stuck.
We were told in class that we could plug in the points and get the equations:
In this case the points are $(0,0),(0,1),(1,0)$.
So if we put the first point we get $u=0,v=0$ or $u=-v,u=v$
If we put in the second point we get
$u=-1,v=1$, $u=-v,u=v+2$
Similarly
$u=1,v=-1$,$u=v,u=v-2$
How do I proceed after this? This is making me even more confused on what values to take and what values not to take.can anyone help me out here with an easy explanation?
 A: given integral =$$\int_0^1(\int_0^{1-x}\frac{(x-y)^4}{(x+y)^4}dy)dx$$Evaluate the inner integral by the change of variable $z=x+y$ (remember that $x$ is constant in the inner integral) which gives $$\int_x^1\frac{(2x-z)^4}{z^4}dz$$ Expand the numerator, integrate term by term , plug in the limits  $z=1\text{ and }z=x$and subtract.You get  an expression in $x$ which you can integrate from 0 to 1.(an improper integral:  integrate from $a$ to 1 and take the limit as $a \rightarrow 0.$)
A: The invertible linear map $\Phi(x,y)=(u,v)=(x-y,x+y)$ transforms lines into lines and therefore if $D=\{(x,y): x\geq 0, y\geq 0, x+y\leq 1\}$, i.e. the triangle with vertices $(0,0)$, $(1,0)$, $(0,1)$, then $\Phi(D)$ is the triangle with vertices $\Phi(0,0)=(0,0)$, $\Phi(1,0)=(1,1)$, $\Phi(0,1)=(-1,1)$.

Hence
$$\begin{align}
\iint_D \left(\frac{x-y}{x+y}\right)^4\,dx dy
&=\iint_{\Phi(D)} \left(\frac{u}{v}\right)^4 |\det(J_{\Phi^{-1}}(u,v)|\,du dv\\
&=\int_{v=0}^1\int_{u=-v}^v\frac{u^4}{v^4}\cdot \frac{1}{2}\,du dv\\
&=\frac{1}{2}\int_{v=0}^1\frac{1}{v^4}\left[\frac{u^5}{5}\right]_{-v}^v\,dv\\
&=\frac{1}{5}\int_{v=0}^1 v\,dv=\frac{1}{10}.
\end{align}$$
P.S. The integrand function $f(x,y)=\left(\frac{x-y}{x+y}\right)^4$ is not defined at $(0,0)\in D$, but it is continuous and bounded by $1$ in $D\setminus\{(0,0)\}$. Therefore the integral of $f$ over $D$ is finite.
A: No change of variable seems necessary.I'll use Sage to explain.
Define the function :
$$f(x,y)=\left(\frac{x-y}{x+y}\right)^4$$
sage: f(x,y)=((x-y)/(x+y))^4

Find an antiderivative of this function with respect to $y$ :
sage: A1=integrate(f(x, y), y) ; A1
-8*x*log(x + y) + y - 8/3*(5*x^4 + 12*x^3*y + 9*x^2*y^2)/(x^3 + 3*x^2*y + 3*x*y^2 + y^3)

$$\int\left(\frac{x-y}{x+y}\right)^4\,dy\,=\,-8x\log(x + y) + y - \frac{8}{3}\frac{5x^4 + 12x^3y + 9x^2y^2}{x^3 + 3x^2y + 3xy2 + y^3}$$
This function is defined and continuous on the square $(0,\ 1],(0,\ 1]$ except in $(0,\ 0)$. Its limits give us the defined integral $\displaystyle\int_0^{1-x}f(x,y)\,dy$ :
sage: D1 = A1.limit(y=1-x)-A1.limit(y=0) ; D1
-16/3*x^4 + 16*x^3 - 24*x^2 + 8*x*log(x) + 37/3*x + 1

$$\int_0^{1-x}\left(\frac{x-y}{x+y}\right)^4\,dy\,=\,-\frac{16}{3}x^4 + 16x^3 - 24x^2 + 8x\log(x) + \frac{37}{3}x + 1$$
which is a function of $x$, defined and continuous on $\mathbb{R}^{*+}$, which has an antiderivative :
sage: A2 = integrate(D1, x) ; A2
-16/15*x^5 + 4*x^4 - 8*x^3 + 4*x^2*log(x) + 25/6*x^2 + x

$$\int\frac{16}{3}x^4 + 16x^3 - 24x^2 + 8x\log(x) + \frac{37}{3}x + 1\,dx\,=\,-\frac{16}{15}x^5 + 4x^4 - 8x^3 + 4x^2\log(x) + \frac{25}{6}x^2 + x$$
continuous on $\mathbb{R}$, whose limits give us the sought integral :
sage: D2 = A2.limit(x=1) - A2.limit(x=0) ; D2
1/10

BTW :
sage: integrate(integrate(f(x, y), y, 0, 1-x, algorithm="sympy"), x, 0, 1)
1/10

(The algorithm="sympy" parameter avoids Sage's default integrator (Maxima), whic is nosy about the value of $x$).
