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Two iterated improper integrals over the unit square are $π / 4$ and $- π / 4$

Example

On $(0, 1)^{2}$ , let

$f (x, y) : = \frac{x^{2} - y^{2}}{(x^{2} + y^{2})^{2}} .$

Both iterated improper integrals exist, but

$\int_{0}^{1} (\int_{0}^{1} f (x, y) d y) d x = \frac{π}{4}, \int_{0}^{1} (\int_{0}^{1} f (x, y) d x) d y = - \frac{π}{4} .$

Moreover, two compact Jordan exhaustions give these different limiting values, so $f$ has no exhaustion-independent improper double integral.

Facts & Assumptions

Given: The function $f$ on the open unit square.

[L1]

The principal inverse tangent satisfies $(\arctan u)^{'} = 1 / (1 + u^{2})$ and $\arctan u = \int_{0}^{u} (1 + t^{2})^{- 1} d t$ (Principal arctangent: derivative, integral, power series, and the Gregory–Leibniz series).

[L2]

A locally Riemann-integrable signed function is improperly integrable precisely when the improper integral of its absolute value is finite (Improper multiple integrals and absolute convergence on open sets).

[L3]

Absolute improper convergence makes every compact Jordan exhaustion converge to the same signed value (Absolute convergence makes signed improper multiple integrals independent of exhaustion).

[L4]

If $G^{'} = h$ on a compact interval and $h$ is integrable, then $\int h$ is the endpoint difference of $G$ (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ ).

Verification

technique · direct

1.1algebra

Direct differentiation gives $f (x, y) = \partial_{y} (y / (x^{2} + y^{2})) = - \partial_{x} (x / (x^{2} + y^{2}))$ .

2.1step 1.1L1L4

Integrating the first identity in $y$ gives $1 / (1 + x^{2})$ , and integrating the second in $x$ gives $- 1 / (1 + y^{2})$ ; [L4] and [L1] therefore give the two iterated values $π / 4$ and $- π / 4$ .

3.1step 1.1L1L2L3∎

On $[a, u] \times [b, v] \subset (0, 1)^{2}$ , step 1.1 gives the double integral $\arctan (u / v) - \arctan (a / v) - \arctan (u / b) + \arctan (a / b)$ . Taking $u_{j} = v_{j} = 1 - 1 / (j + 2)$ and either $(a_{j}, b_{j}) = ((j + 2)^{- 2}, (j + 2)^{- 1})$ or the swapped pair produces nested compact Jordan exhaustions with limits $- π / 4$ and $π / 4$ . By [L3], and hence by [L2], no exhaustion-independent signed improper integral exists.

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Two iterated improper integrals over the unit square are π/4π/4 and −π/4−π/4