To evaluate a double integral $I=\int\int \frac{x-y}{x+y} dx dy$ over the rectangle $[0,1;0,1]$. To evaluate the Double Integral
$$I=\int\int \frac{x-y}{x+y} dx dy$$
over the rectangle $[0,1;0,1]$.
My Observation:-

*

*The function $f(x,y)=\frac{x-y}{x+y}$ is discontinuous only at $(0,0)$. Hence the double integral exist.


*for $y\neq 0$, $\phi(y)=\int^1_0 f(x,y) dx=1+2y log \frac{y}{1+y}$, showing that $\phi(0)$ does not exist. Hence the double intgral may not be equal to $J=\int^1_0\int^1_0 f(x,y)dx dy$.


*if $\phi(0)$, exist, then $I=J$.


*Similarly $\psi(x)=\int^1_0 f(x,y) dy$ does not exist for $x=0$.
So how to evaluate the double integral. Any help is highly appreciated.
 A: Let us check integrability first. Fix some $\varepsilon >0$, then:
$$
\begin{aligned}
&\int_\varepsilon^1
\int_0^1
\left|\frac{x-y}{x+y}\right|\; dx\;dy
\\
&\qquad
=
\int_\varepsilon^1
\int_0^x
\frac{x-y}{x+y}\; dx\;dy
-
\int_\varepsilon^1
\int_x^1
\frac{x-y}{x+y}\; dx\;dy
\\
&\qquad
=
\int_\varepsilon^1 dx\; \Big[\ 2x\log(x+y) -y\ \Big]_{y=0}^{y=x}
-
\int_\varepsilon^1 dx\; \Big[\ 2x\log(x+y) -y\ \Big]_{y=x}^{y=1}
\\
&\qquad
=
\int_\varepsilon^1 \Big(\ 2x\log\frac{x+x}{x+0} -(x-0)\ \Big)\; dx
-
\int_\varepsilon^1 \Big(\ 2x\log\frac{x+1}{x+x} -(1-x)\ \Big)\; dx
\\
&\qquad
\nearrow 
\left(\log2-\frac 12\right)
-
\left(-\log2+\frac 12\right)
\qquad\text{ for }\varepsilon\searrow 0\ .
\end{aligned}
$$

Now in order to see that the integral is zero, use the substitution of variables $(X,Y)=(y,x)$, or copy the above explicit computation starting without the modulus, and using a plus sign "in the middle" between the integrals.
$\square$
A: Let $R=[0,1]\times[0,1]$ the rectangular region. Define the change of variables,
$$u=x-y,\quad v=x+y,$$
then,  $u\in [-1,1]$ and $v\in [0,2]$ since $0\leqslant x\leqslant 1$ and $0\leqslant y\leqslant 1$. By the change of variables theorem, we have $\iint_{R}f(x,y)\,dA=\iint_{R^{*}}f(u,v)\left|J\right|dA^{*}$, where the determinant of Jacobian matrix $J$ is given by $\det\begin{pmatrix}x_{u}&x_v\\y_u&y_v\end{pmatrix}$ in this case $|J|=1/2$. Hence, $\int_{0}^{1}\int_{0}^{1}\frac{x-y}{x+y}dxdy=\frac{1}{2}\int_{0}^{2}\frac{1}{v}\left(\int_{-1}^{1}udu\right)dv=0$.
