Does changing the order of double integration (both integral limits are constants) alter the final answer?

Question

I have studied that changing the order of double integration will not change the answer if both the limits of integration are constants. But this function is not agreeing with what I have studied: $$1)\int_0^1\left(\int_0^1{\frac{x-y}{(x+y)^3}dy}\right)dx$$ $$2)\int_0^1\left(\int_0^1{\frac{x-y}{(x+y)^3}dx}\right)dy$$ The answer to the first integral is 0.5 and that of the second integral is -0.5 respectively.

Can anyone please explain why is this so?

Answer 1

In general you cannot switch the order of integration without additional constraints. These are typically given by Fubini's theorem. In particular the example you've given does not converge absolutely so switching the order changes the answer.

Answer 2

We have \begin{align} &\int_0^1\left(\int_0^1{\frac{x-y}{(x+y)^3}dx}\right)dy\\ &=\int_0^1\left(\int_0^1{\frac{x+y-2y}{(x+y)^3}dx}\right)dy\\ &=\int_0^1\left[\int_0^1\left(\frac{1}{(x+y)^2}-\frac{2y}{(x+y)^3}\right)dx\right]dy\\ &=\int_0^1\left[\frac{-1}{x+y}+\frac{2y}{2(x+y)^2}\right]_{x=0}^1dy\\ &=\int_0^1\left[\frac{-1}{(1+y)^2}\right]dy\\ &=\left[\frac{1}{1+y}\right]_0^1=-\frac12. \end{align} Similarly, you can show that \begin{align} \int_0^1\left(\int_0^1{\frac{x-y}{(x+y)^3}dy}\right)dx=\frac12. \end{align}