1
$\begingroup$

I'm reading Buck's advanced calculus. It says for improper integral of higher dimensions, conditional convergence is impossible, i.e., $\int\int_D f$ cannot exist without $\int\int_D|f|$ existing too.
Then book only gives a sketch of proof as follow.
Let $f_1=(|f|+f)/2$ and $f_2=(|f|-f)/2$. We may assume that the integrals $\int\int_Df_i$ are each divergent. Since $f_1f_2=0$, so that the sets where $f_1$ and $f_2$ are positive are disjoint. It is then possible to choose an expanding sequence of closed rectangles $\{D_n\}$ which favour $f_1$ over $f_2$, so that $\int\int_{D_n} f_1$ diverges faster than $\int\int_{D_n} f_2$, with the result that $\int\int_{D_n} f$, which is their different, also diverge.
But it feels like a almost exactly same proof can be used to show that single improper integral can not be conditional convergent too, but single integral can be convergent without being absolute convergent. For example, $\int^\infty_1 x^{-1}\sin x$ is conditional convergent but not absolutely convergent.
So what is the essential difference between single integral and double integral which makes the conditional convergence for double integral impossible? Thanks.

$\endgroup$
1
$\begingroup$

Suppose $f : D \subset \mathbb{R}^d \to \mathbb{R}$ is Riemann integrable on every compact rectifiable subset of $D$. The multiple improper integral is defined generally as

$$\int_D f = \lim_{n \to \infty} \int_{D_n}f,$$

where $(D_n)$ is a sequence of compact rectifiable sets such that $D_n \subset \text{int } D_{n+1}$ and $\cup_{n=1}^\infty D_n = D$. The improper integral is well-defined if the limit does not depend on the choice for $(D_n)$.

Under such a definition, it must hold that the improper integral of $f$ over $D$ exists if and only if the improper integral of $|f|$ exists over $D$.

In one-dimension ($d = 1$), the improper integral can be conditionally convergent when defined specifically as a limit of integrals over nested intervals , such as $D_n = [0,n]$ where

$$\int_0^\infty \frac{\sin x}{x} \, dx := \lim_{n \to \infty}\int_0^n \frac{\sin x}{x} \, dx = \frac{\pi}{2}$$

However, even in one-dimension, the more general definition of the improper integral precludes conditional convergence.

For example, consider the following sequence $D_n \subset [0,\infty)$ where each set is a finite union of intervals with gaps,

$$D_n = [0, (2n-1)\pi] \cup \bigcup_{k=n}^{2n}[2k\pi,(2k+1)\pi ]$$

It is easy to show that $D_n \subset D_{n+1}$ for all $n$. Also, for any $c > 0$, there exists $n$ such that $(2n-1)\pi > c$ and $[0,c] \subset D_n$, and this implies $\cup_n D_n = [0,\infty)$.

The integral over $D_n$ is

$$\int_{D_n}\frac{\sin x }{x} \, dx = \int_0^{(2n-1)\pi }\frac{\sin x }{x} \, dx + \sum_{k=n}^{2n} \int_{2k \pi}^{(2k+1) \pi } \frac{\sin x} {x} \, dx,$$

which can be shown to converge to a value greater than $\pi/2 + \log 2 /\pi$.

The first integral on the right-hand side converges to $\pi/2$ and, since $\sin x \geqslant 0$ for $x \in [2k \pi,(2k+1) \pi ]$, it follows that

$$\int_{2k \pi}^{(2k+1) \pi } \frac{\sin x} {x} \, dx > \frac{1}{(2k+1)\pi }\int_{2k \pi}^{(2k+1) \pi } \sin x \, dx = \frac{2}{(2k+1)\pi } > \frac{1}{\pi}\frac{1}{k+1}$$

Thus,

$$\limsup_{n \to \infty}\int_{D_n} \frac{\sin x }{x} \, dx > \frac{\pi}{2} + \lim_{n \to \infty} \frac{1}{\pi} \sum_{k = n}^{2n}\frac{1}{k+1} = \frac{\pi}{2}+ \frac{\log 2}{\pi}$$

There can be no unique value of the limit of the integral over $D_n$ for every choice of sequence $(D_n)$.

$\endgroup$
2
  • $\begingroup$ Thanks! It really makes things clear! $\endgroup$ – Cathy Jul 13 '20 at 19:36
  • $\begingroup$ @Cathy: You’re welcome. $\endgroup$ – RRL Jul 13 '20 at 23:00
0
$\begingroup$

I don't know whether it is always possible to choose such a sequence of rectangles $D_n$.

In my view the difference with the one-dimensional situation is the following: For integrals $\int_0^\infty$ there is just one reasonable way to pass to the limit, namely looking at $\lim_{b\to\infty}\int_0^b$. The improper integral then is equal to this limit. But already for integrals $\int_{-\infty}^\infty$ there is the question whether you should consider the limit $\lim_{N\to\infty}\int_{-N}^N$ (sometimes called principal value) or the two independent variables limit $\lim_{a\to-\infty, \>b\to\infty}\int_a^b$.

Now in the multivariable case there are infinitely many ways to define increasing sequences $(A_n)_{n\geq0}$ of subsets that exhaust ${\mathbb R}^2$, and depending on the case you may obtain different limits $\lim_{n\to\infty}\int_{A_n}$. Of course you could say: I definitely consider only principal values, where the integral is over balls $B_r\subset{\mathbb R}^2$. In such a setting you could obtain convergent improper integrals for certain functions with divergent $\int_{{\mathbb R}^2}|f(x,y)|\>{\rm d}(x,y)$. This might be fine in geometric settings, but is of no value when the variables $x$, $y$ are of different semantical types.

To sum it up: In ${\mathbb R}^d$, $\>d\geq2$, there is no single "canonical" type of exhausting space in order to define a "universal" improper integral. In the Lebesgue world one requires (also in the one-dimensional case) that $\int_{{\mathbb R}^d}|f(x)|\>{\rm d}(x)$ is finite.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.