{
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  "body": "\n## Counterexample\n\nOne could guess that the two [[iterated integrals|0P4WX7Y502]] of a function over a square are always equal when both exist.\nThis is false.\nLet $f(x, y) = \\dfrac{x^2 - y^2}{(x^2 + y^2)^2}$ on $(0, 1] \\times (0, 1]$. Then\n$$\\int_0^1 \\int_0^1 f(x, y) \\, dy \\, dx = \\frac{\\pi}{4}, \\qquad \\int_0^1 \\int_0^1 f(x, y) \\, dx \\, dy = -\\frac{\\pi}{4}.$$\n\n## Remarks\n\nFor fixed $x \u003e 0$, $\\partial_y \\big( y / (x^2 + y^2) \\big) = f(x, y)$. 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 $-\\pi / 4$.\nThe function is not integrable on the square: $\\int_0^1 \\int_0^1 |f| \\, dy \\, dx = \\infty$. So [[Fubini's theorem|0R0T1VWH5X]] does not apply.\nA 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 $1$.\n\n## Explanations\n\n### L1 human\n\nNear the corner $(0, 0)$ the function has large positive values and large negative values.\nThe two orders of integration cancel these parts in different ways, so they give different answers.\n",
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        "locator": "Failure of Fubini's theorem for non-integrable functions",
        "url": "https://en.wikipedia.org/wiki/Fubini%27s_theorem"
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        "locator": "Lack of commutativity",
        "url": "https://en.wikipedia.org/wiki/Iterated_integral"
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  "msc": [
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  "name": "Function for which the iterated integrals differ",
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