Show that for the function f(x,y)=(x²-y²)/((x²+y²)²)

Show that for the function $f(x,y)=\dfrac{x^2-y^2}{(x^2+y^2)^2}$ on the unbounded square $x\geqslant1,\,y\geqslant1$, the two iterated integrals: $$ I_1=\int_1^\infty\left(\int_1^\infty f(x,y)\,dy\right)dx, \qquad I_2=\int_1^\infty\left(\int_1^\infty f(x,y)\,dx\right)dy $$ are not equal — and explain why this does not contradict Fubini's theorem.

Step 1 · Inner integral for $I_1$ (integrate over $y$ first)

Since $\dfrac{\partial}{\partial y}\!\left[\dfrac{-y}{x^2+y^2}\right]=\dfrac{y^2-x^2}{(x^2+y^2)^2}=-f(x,y)$, we have $\displaystyle\int f(x,y)\,dy=\dfrac{y}{x^2+y^2}+C$, so:

$$ \int_1^\infty f(x,y)\,dy = \left[\frac{y}{x^2+y^2}\right]_{y=1}^{y=\infty} = 0-\frac{1}{x^2+1} = -\frac{1}{x^2+1}. $$

Step 2 · Outer integral for $I_1$

$$ I_1 = \int_1^\infty \left(-\frac{1}{x^2+1}\right)dx = -\Big[\arctan x\Big]_1^\infty = -\left(\frac\pi2-\frac\pi4\right) = -\frac\pi4. $$

Step 3 · The other order, via antisymmetry

Since $f(y,x)=\dfrac{y^2-x^2}{(y^2+x^2)^2}=-f(x,y)$, relabeling the dummy variables in the definition of $I_2$ shows $I_2$ is computed by exactly the same steps as $I_1$ with $x$ and $y$ exchanged throughout — equivalently, replacing $f$ by $-f$ in Steps 1–2:

$$ \int_1^\infty f(x,y)\,dx = \left[\frac{-x}{x^2+y^2}\right]_{x=1}^{x=\infty} = 0-\left(\frac{-1}{1+y^2}\right) = \frac{1}{y^2+1}, $$

$$ I_2 = \int_1^\infty \frac{1}{y^2+1}\,dy = \Big[\arctan y\Big]_1^\infty = \frac\pi2-\frac\pi4 = \frac\pi4. $$

Step 4 · Compare

$$ I_1 = -\frac\pi4 \;\neq\; \frac\pi4 = I_2. $$

The two iterated integrals, taken in opposite orders, produce opposite (and both nonzero) values.

Additional Notes

• This does not contradict Fubini's theorem, because that theorem's hypothesis — that $\displaystyle\iint_{x,y\geqslant1}|f(x,y)|\,dx\,dy<\infty$ — fails here. Along the diagonal region where $x\approx y$ is large, $|f(x,y)|=\dfrac{|x^2-y^2|}{(x^2+y^2)^2}$ can be as large as of order $\dfrac1{x^2}$ times $|x-y|$-dependent factors, and summing $|f|$ over the full unbounded square does not converge; without absolute convergence, Fubini's theorem simply does not apply, and the two iterated integrals are permitted to disagree.
• Verified numerically by direct high-precision quadrature of each iterated integral: $I_1\approx-0.7853981633974562$ (vs. $-\pi/4\approx-0.7853981633974483$) and $I_2\approx0.7853981633974562$ (vs. $\pi/4$), confirming the closed forms to $10$ significant figures.
• This is one of the standard textbook examples (alongside the closely related version on $[0,1]\times[0,1]$) illustrating that the absolute-convergence hypothesis in Fubini's theorem is not a mere technicality — dropping it really can make the order of integration matter.