Is there any example of a function $f:[a,b]\times[c,d]\to \mathbb R$ so that $\int_a^b\int_c^d f(x,y)\,dy\,dx$ and $\int_c^d\int_a^b f(x,y)\,dx\,dy$ exist and are equal but $\int\int f(x,y)\,dy\,dx$ does not exist in $[a,b]\times[c,d]$?
I've been trying to find an example but I have nothing so far.
The function $f:[-1,1]^2 \to \mathbb{R}$ where
$$f(x,y) = \begin{cases}\frac{xy}{(x^2 +y^2)^2}, \,\,\,\ x^2 + y^2 > 0\\ 0, \,\,\,\, x^2 + y^2 = 0 \end{cases}$$
is not integrable, but
$$\int_{-1}^1 \left(\int_{-1}^1 f(x,y) \, dx \right) \, dy = \int_{-1}^1 \left(\int_{-1}^1 f(x,y) \, dy \right) \, dx = 0.$$
Also consider
$$f(x,y) = \begin{cases}\frac{xy^2}{(x^2 +y^2)^2}, \,\,\,\ x^2 + y^2 > 0\\ 0, \,\,\,\, x^2 + y^2 = 0 \end{cases}$$
where
$$\int_{0}^1 \left(\int_{0}^1 f(x,y) \, dx \right) \, dy = \int_{0}^1 \left(\int_{0}^1 f(x,y) \, dy \right) \, dx = \frac{\pi}{8},$$
but $\int_{[0,1]^2}f $ exists only as an improper integral.
Leaving details to you.