The iterated integral

$\backslash int\_0^1\backslash int\_0^1\; \backslash frac\{x^2-y^2\}\{(x^2+y^2)^2\}\backslash ,dy\backslash ,dx$

does not converge absolutely, i.e. the integral of the absolute value is not finite:

$\backslash int\_0^1\backslash int\_0^1$

\left|\frac{x^2-y^2}{(x^2+y^2)^2}\right|\,dy\,dx=\infty.

Fubini's theorem tells us that if the integral of the absolute value is finite, then the order of integration does not matter; if we integrate first with respect to x and then with respect to y, we get the same result as if we integrate first with respect to y and then with respect to x. The assumption that the integral of the absolute value is finite is "Lebesgue integrability". That the assumption of Lebesgue integrability in Fubini's theorem cannot be dropped can be seen by examining this particular iterated integral. Clearly putting "dx dy" in place of "dy dx" has the effect of multiplying the value of the integral by −1 because of the "antisymmetry" of the function being integrated. Therefore, unless the value of the integral is zero, putting "dx dy" in place of "dy dx" actually changes the value of the integral. That is indeed what happens in this case.

How to prove that $\backslash int\_0^1\backslash int\_0^1\; \backslash left|\backslash frac\{x^2-y^2\}\{(x^2+y^2)^2\}\backslash right|\backslash ,dx\backslash ,dy$ does not converge

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One way to do this without using Fubini's theorem is as follows:

$\backslash int\_0^1\backslash int\_0^1$

\left|\frac{x^2-y^2}{(x^2+y^2)^2}\right|\,dx\,dy=\int_0^1\left[\int_0^y \frac{y^2-x^2}{(x^2+y^2)^2}\,dx+\int_y^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dx\right]\,dy

$=\backslash int\_0^1\backslash left(\backslash frac\{1\}\{2y\}+\backslash frac\{1\}\{2y\}-\backslash frac\{1\}\{y^2+1\}\backslash right)\backslash ,dy=\backslash int\_0^1\; \backslash frac\{1\}\{y\}\backslash ,dy-\backslash int\_0^1\backslash frac\{1\}\{1+y^2\}\backslash ,dy.$

How to evaluate this integral

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The integral

$\backslash int\_0^1\backslash frac\{x^2-y^2\}\{(x^2+y^2)^2\}\backslash ,dy$

can be evaluated via the trigonometric substitution

$y=x\backslash tan(\backslash theta),$

$dy=x\backslash sec^2(\backslash theta)\backslash ,d\backslash theta,$

$x^2+y^2=x^2+x^2\backslash tan^2(\backslash theta)=x^2(1+\backslash tan^2(\backslash theta))$

=x^2\sec^2(\theta).

The bounds of integration can be found thus:

$0\backslash leq\; y\backslash leq\; 1,$

$0\backslash leq\; x\backslash tan(\backslash theta)\backslash leq\; 1,$

$0\backslash leq\backslash tan(\backslash theta)\backslash leq\; 1/x,$

$0\backslash leq\backslash theta\backslash leq\backslash arctan(1/x).$

The integral then becomes

$\backslash int\_0^\{\backslash arctan(1/x)\}$

\frac{x^2(1-\tan^2(\theta))}{(x^2\sec^2(\theta))^2} x\sec^2(\theta)\,d\theta =\frac{1}{x}\int_0^{\arctan(1/x)} \frac{1-\tan^2(\theta)}{\sec^2(\theta)}\,d\theta

$=\backslash frac\{1\}\{x\}\backslash int\_0^\{\backslash arctan(1/x)\}$

\cos^2(\theta)-\sin^2(\theta)\,d\theta =\frac{1}{x}\int_0^{\arctan(1/x)} \cos(2\theta)\,d\theta =\frac{1}{x}\left[\frac{\sin(2\theta)}{2} \right]_{\theta:=0}^{\theta=\arctan(1/x)}

$=\backslash frac\{1\}\{x\}\backslash left[\backslash sin(\backslash theta)\backslash cos(\backslash theta)$

\right]_{\theta:=0}^{\theta=\arctan(1/x)} =\frac{1}{x}\sin(\arctan(1/x))\cos(\arctan(1/x)).

Now recall the trigonometric identities

$\backslash sin(\backslash arctan(1/x))=\backslash frac\{1\}\{\backslash sqrt\{1+x^2\}\}$

\ \mbox{and}\ \cos(\arctan(1/x))=\frac{x}{\sqrt{1+x^2}}.

The expression above then becomes

$\backslash frac\{1\}\{1+x^2\}.$

This takes care of the "inside" integral with respect to y; now we do the "outside" integral with respect to x:

$\backslash int\_0^1\backslash frac\{1\}\{1+x^2\}\backslash ,dx$

=\left^*_0^1

=\arctan(1)-\arctan(0)=\frac{\pi}{4}.

Thus we have

$\backslash int\_0^1\backslash int\_0^1\backslash frac\{x^2-y^2\}\{(x^2+y^2)^2\}\backslash ,dy\backslash ,dx=\backslash frac\{\backslash pi\}\{4\}$

and

$\backslash int\_0^1\backslash int\_0^1\backslash frac\{x^2-y^2\}\{(x^2+y^2)^2\}\backslash ,dx\backslash ,dy=-\backslash frac\{\backslash pi\}\{4\}.$

Fubini's theorem implies that since these two iterated integrals differ, the integral of the absolute value must be ∞.

The moral of the story

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When

$\backslash int\_a^b\backslash int\_c^d\; \backslash left|f(x,y)\backslash right|\backslash ,dy\backslash ,dx=\backslash infty$

then the two iterated integrals

$\backslash int\_a^b\backslash int\_c^d\; f(x,y)\backslash ,dy\backslash ,dx\backslash \; \backslash mbox\{and\}\backslash \; \backslash int\_c^d\backslash int\_a^b\; f(x,y)\backslash ,dx\backslash ,dy$

may have different finite values.

Multivariate calculus