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

Question

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?

Answer 1

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.

Answer 2

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 $a \rightarrow 0.$)

Answer 3

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 $x$).