<s>[INST] Proving that $\iint\frac{x-y}{(x+y)^3}\,dx\,dy$ does not exist over $0 \leq x \leq 1 , 0 \leq y \leq 1$ To prove that double integral does not exist:     

$$\iint\frac{x-y}{(x+y)^2}\,dx\,dy$$ over region $0 \leq x \leq 1 , 0 \leq y \leq 1$.

I put $x - y = u$ and $x+y = v$ and I got integral as 
$$\iint \frac{u}{v^3}\,du\,dv$$ 
Limits of $u$ are from $2-v \leq u\leq v-2$ and $v$ are from $0$ to $1$. I am not sure about my new limits and how to prove it further.
EDIT : The correct expression is $\iint\frac{x-y}{(x+y)^3}\,dx\,dy$
 [/INST] Given Integral $$ \begin{align} I &= \int^1_0 \int^1_0 \cfrac{x-y}{(x+y)^2} dx dy \\
                                  & = \int^1_0 \int^1_0 \cfrac{x+y-2y}{(x+y)^2} dx dy \\ & = \int^1_0 \int^1_0 \cfrac{1}{(x+y)} dx dy - 2 \int^1_0 \int^1_0 \cfrac{y}{(x+y)^2} dx dy \\ & = \int^1_0 \ln(x+y) \bigg|_0^1 dy + 2 \int^1_0 \cfrac{y}{x+y} \bigg |^1_0 dy  \\ & = \int^1_0\ln (y+1) dy - \int^1_0\ln(y) dy   - 2 \int^1_0 \cfrac{dy}{y+1} \\& = (y+1)(\ln(y+1)-1  )\bigg|_0^1 - y(\ln(y)-1)  \bigg|_0^1 - 2 \ln(y+1) \bigg |^1_0 \end{align}$$
Which is easy to evaluate, and I think the only slight problem would be in evaluating $y \ln(y) $ at $x = 0$, which is just $0$.
I don't see any problem with this.

After changing the question
Given integral:
$$ \begin{align} I &= \int^1_0 \int^1_0 \cfrac{x-y}{(x+y)^3} dx dy \\
                                  & = \int^1_0 \int^1_0 \cfrac{x+y-2y}{(x+y)^3} dx dy  \\& = \int^1_0 \int^1_0 \cfrac{dx dy}{(x+y)^2} - 2 \int^1_0 \int_0^1 \cfrac{y}{(x+y)^3} dx dy \\& = -\int^1_0  \cfrac{ dy}{(x+y)} \bigg |^1_0 + \cfrac{2}{2} \int^1_0  \cfrac{y}{(x+y)^2} \bigg |^1_0 dy  \\& = - \int^1_0 \bigg( \cfrac{1}{y+1} - \cfrac{1}{y} \bigg) dy + \int^1_0 \bigg( \cfrac{y}{(1+y)^2} - \cfrac{1}{y} \bigg ) dy \\ & = - \int^1_0 \cfrac{dy}{(1+y)^2}  \\& = {- \cfrac{1}{2}}\end{align}$$
But $$ \int^1_0 \int^1_0 \cfrac{x-y}{(x+y)^3} dy dx = \cfrac{1}{2} $$
Then from converse of Fubini's theorem this integral doesn't exists.
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