The reason that conditionally convergent integral produce different results

Question

A well known result is that for \begin{equation} f(x,\,y)=\frac{x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}, \end{equation} one obtains \begin{equation} \int_{1}^{\infty}\int_{1}^{\infty}f(x,y)dxdy=\frac{\pi}{4}\qquad\int_{1}^{\infty}\int_{1}^{\infty}f(x,y)dydx=-\frac{\pi}{4}. \end{equation} What I would like to understand better is what cause that difference between the results. Noticeably, for any integration involving only finite values of $\Lambda$, \begin{equation} \int_{1}^{\Lambda}\int_{1}^{\Lambda}f(x,y)dxdy=\int_{1}^{\Lambda}\int_{1}^{\Lambda}f(x,y)dydx=0, \end{equation} \begin{equation} \int_{1}^{\Lambda}\int_{\Lambda}^{\infty}f(x,y)dxdy+\int_{\Lambda}^{\infty}\int_{1}^{\Lambda}f(x,y)dxdy=0. \end{equation} So it seems like the entire contribution comes from treating the region "beyond the finite". By writting \begin{equation} f(x,y)=-\frac{\partial}{\partial x}\frac{\partial}{\partial y}\arctan\left(\frac{y}{x}\right)=\frac{\partial}{\partial y}\frac{\partial}{\partial x}\arctan\left(\frac{x}{y}\right), \end{equation} we indeed realize that the correct result is reproduced by the following assignment \begin{equation}\begin{split} &\int_{\Lambda}^{\infty}\int_{\Lambda}^{\infty}f(x,y)dxdy=\lim_{(x,y)\rightarrow(\infty,\infty)}\arctan\left(\frac{x}{y}\right)=\frac{\pi}{4},\\ &\int_{\Lambda}^{\infty}\int_{\Lambda}^{\infty}f(x,y)dydx=-\lim_{(x,y)\rightarrow(\infty,\infty)}\arctan\left(\frac{y}{x}\right)=-\frac{\pi}{4}. \end{split}\end{equation} What I am not sure if this is if this treatment is justified (although, it does work in other examples as well). In addition, the Clairaut's theorem seems not to be valid in the asymptotic region (otherwise the results must have been equal), why is it so?

Answer 1

Since the integrand is continuous, it is absolutely integrable over any finite region excluding the point $(0,0)$. Fubini's theorem guarantees that the iterated integrals must be equal to one another as well as to the double integral. We can interpret the double integral here as either the Lebesgue integral or the (multivariate) Riemann integral. That is,

$$\int_{[1,A]\times[1,B]}\frac{x^2-y^2}{(x^2+y^2)}d(x,y)= \int_1^A\left(\int_1^B \frac{x^2-y^2}{(x^2+y^2)}\, dy\right)\,dx=\int_1^B\left(\int_1^A \frac{x^2-y^2}{(x^2+y^2)}\, dx\right)\,dy$$

Now consider the iterated improper integrals

$$I_1=\int_1^\infty\left(\int_1^\infty \frac{x^2-y^2}{(x^2+y^2)}\, dy\right)\,dx, \quad I_2=\int_1^\infty\left(\int_1^\infty \frac{x^2-y^2}{(x^2+y^2)}\, dx\right)\,dy$$

The correct meaning of $I_1$ and $I_2$ is an iterated limit of integrals over finite intervals $1 \leqslant y \leqslant B$ and $1 \leqslant x \leqslant A$, where $A$ and $B$ tend independently to $+\infty$, that is

$$\tag{1}\begin{align}I_1 &= \lim_{A\to +\infty}\lim_{B \to +\infty}\int_1^A\left(\int_1^B \frac{x^2-y^2}{(x^2+y^2)}\, dy\right)\,dx \\&= \lim_{A\to +\infty}\lim_{B \to +\infty}\int_1^A\left(\int_1^B \frac{\partial}{\partial y}\frac{\partial}{\partial x}\arctan\left(\frac{x}{y}\right)\, dy\right)\,dx\\&= \lim_{A\to +\infty}\lim_{B \to +\infty}\left(\arctan\frac{A}{B}- \arctan\frac{1}{B} - \arctan \frac{A}{1} + \arctan \frac{1}{1}\right)\\ &= \lim_{A\to \infty}\left(0-0 - \arctan A + \frac{\pi}{4}\right)= -\frac{\pi}{2}+\frac{\pi}{4}= - \frac{\pi}{4}\end{align}$$

Similarly,

$$\tag{2}\begin{align}I_2 &= \lim_{B\to +\infty}\lim_{A \to +\infty}\int_1^B\left(\int_1^A -\frac{\partial}{\partial x}\frac{\partial}{\partial y}\arctan\left(\frac{y}{x}\right)\, dx\right)\,dy\\&= -\lim_{B\to +\infty}\lim_{A \to +\infty}\left(\arctan\frac{B}{A}- \arctan\frac{1}{A} - \arctan \frac{B}{1} + \arctan\frac{1}{1}\right)\\ &= -\lim_{B\to \infty}\left(0-0 - \arctan B + \frac{\pi}{4}\right)= \frac{\pi}{2} -\frac{\pi}{4}= \frac{\pi}{4}\end{align}$$

When you choose a common upper limit $A= B= \Lambda$, then, by antisymmetry across the line $y=x$, the contribution from the region $\{(x,y): 1\leqslant x < y \leqslant \Lambda\}$ cancels the contribution from the region $\{(x,y): 1\leqslant y < x \leqslant \Lambda\}$ and we have

$$\int_1^\Lambda\left(\int_1^\Lambda \frac{x^2-y^2}{(x^2+y^2)}\, dy\right)\,dx=\int_1^\Lambda\left(\int_1^A \frac{x^2-y^2}{(x^2+y^2)}\, dx\right)\,dy= 0$$

We can also see this by enforcing $A = B$ in (1) and (2).

In that case, we get

$$\lim_{\Lambda \to +\infty}\int_1^\Lambda\left(\int_1^\Lambda \frac{x^2-y^2}{(x^2+y^2)}\, dy\right)\,dx=\lim_{\Lambda \to +\infty}\int_1^\Lambda\left(\int_1^A \frac{x^2-y^2}{(x^2+y^2)}\, dx\right)\,dy=0$$

Analogy to a double sequence

This is a common problem that arises with double sequences $(x_{mn})$ where the double limit as $(m,n) \to (\infty,\infty)$ fails to exist, but we can have

$$\lim_{m \to \infty} x_{mm} \neq \lim_{m \to \infty} \lim_{n \to \infty}x_{mn} \neq \lim_{n \to \infty} \lim_{m \to \infty}x_{mn}$$

For example, take $x_{mn} = m/(m+n)$. In this case, we have

$$\lim_{m \to \infty} x_{mm} = \lim_{m \to \infty} \frac{m}{m+m} = \frac{1}{2}, \\\lim_{m \to \infty} \lim_{n \to \infty} \frac{m}{m+n} = 0, \\ \lim_{n \to \infty} \lim_{m \to \infty} \frac{m}{m+n}=1$$

However, the double limit does not exist in the sense that there is no number $L$ such that for every $\epsilon > 0$ there exists $N$ such that $|x_{mn} - L| < \epsilon$ for all $m,n > N$.

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Note that the result,

$$\lim_{\Lambda \to +\infty}\int_1^\Lambda\left(\int_1^\Lambda \frac{x^2-y^2}{(x^2+y^2)}\, dy\right)\,dx=\lim_{\Lambda \to +\infty}\int_1^\Lambda\left(\int_1^A \frac{x^2-y^2}{(x^2+y^2)}\, dx\right)\,dy=0,$$

seems to suggest, paradoxically, that the double integral over the infinite region $[1,\infty)\times [1,\infty)$ is zero. But we have to be careful in how we define the double (as opposed to the iterated) integral over an infinite region. In this case, the integrand is not absolutely integrable over the infinite region. Neither the Lebesgue integral nor the multivariate improper Riemann integral of the absolute value of the integrand is finite. Fubini's theorem does not hold, but the iterated improper integrals exist and take different values.

The double improper Riemann integral over an infinite region such as $[0,\infty)^2$ is typically defined as

$$\int_{[1,\infty)^2}f(x,y) \,d(x,y)= \lim_{n\to \infty}\int_{A_n}f(x,y)\, d(x,y), $$

where $(A_n)$ is an increasing sequence of compact, rectifiable sets such that $A_1 \subset A_2 \subset \ldots$ and $[0,\infty)^2 = \cup_{n=1}^\infty A_n$. That limit can only be uniquely determined independent of the particular sequence $(A_n)$ when absolute integrability holds.