Function for which the iterated integrals differ

Counterexample

One could guess that the two iterated integrals of a function over a square are always equal when both exist. This is false. Let $f(x, y) = \dfrac{x^2 - y^2}{(x^2 + y^2)^2}$ on $(0, 1] \times (0, 1]$. Then $$\int_0^1 \int_0^1 f(x, y) \, dy \, dx = \frac{\pi}{4}, \qquad \int_0^1 \int_0^1 f(x, y) \, dx \, dy = -\frac{\pi}{4}.$$

Remarks

For fixed $x > 0$, $\partial_y \big( y / (x^2 + y^2) \big) = f(x, y)$. So the inner integral is $1/(1 + x^2)$, and its integral over $[0, 1]$ is $\pi / 4$. Since $f(y, x) = -f(x, y)$, the other order gives $-\pi / 4$. The function is not integrable on the square: $\int_0^1 \int_0^1 |f| \, dy \, dx = \infty$. So Fubini's theorem does not apply. A discrete example is $a_{xy} = 1$ if $x = y$, $a_{xy} = -1$ if $x = y + 1$, and $a_{xy} = 0$ otherwise, on the positive integers. The two iterated sums are $0$ and $1$.

Explanations

L1 human

Near the corner $(0, 0)$ the function has large positive values and large negative values. The two orders of integration cancel these parts in different ways, so they give different answers.