How to compute $\int_{0}^{1}\int_{0}^{1} \frac{x-y}{(x+y)^3} dxdy$

How can I change the variable in $$\int_{0}^{1}\int_{0}^{1} \frac{x-y}{(x+y)^3} dxdy$$ to computer the value

I tried to let $u=x+y$ and $v=x-y$

But I have no idea for next step

Can you give me some suggestion,thank you!!!

Solution 1:

The integral is divergent in the origin (x,y) = (0,0) and hence the expression simply makes no sense without additional prescriptions about how to circumvent the divergence.

We shall present one possible regularization and show that the so defined integral can assume any real value we want (including infinity).

To begin with, let us exclude the origin introducing two small positive parameters $\epsilon $ and $\delta$:

$$f(\epsilon ,\delta )=\int _{\epsilon }^1\int _{\delta }^1\frac{x-y}{(x+y)^3}dydx$$

The integral is now well defined and the result is independent of the order of integration

$$\text{f1}=\frac{\delta (\epsilon -1)}{(\delta +1) (\delta +\epsilon )}+\frac{1}{\epsilon +1}-\frac{1}{2}$$

We now want to let $\epsilon $ and $\delta$ go to zero.
There are many possibilities to do that.

1) sequential

First $\epsilon \to 0$, then $\delta \to 0$ gives $-\frac{1}{2}$

First $\delta \to 0$, then $\epsilon \to 0$ gives $+\frac{1}{2}$

Thus the result depends on the order.

2) Simultaneously, on a staight line $\delta =q \epsilon$

The limit $\epsilon \to 0$ is

$$f2 = \frac{1-q}{2 q+2}$$

which can assume, by an appropriate choice of q, any real value. The choice q = 1 is normally called principal value and is equal to zero.

Solution 2:

Hint:

$$\frac{x-y}{(x+y)^3} = \frac{\partial}{\partial x} \left(\frac{-x}{(x+y)^2}\right)$$

Then you should easily find for the iterated integral

$$\int_{0}^{1}\int_{0}^{1} \frac{x-y}{(x+y)^3} \, dx \, dy = -1/2$$

Note that switching the order of integration gives a value of $+1/2$ by antisymmetry of the integrand.