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We consider definite and convergent integrals where the integration (Lebesgue or Riemann integral) $$\int_{S} \ldots d\mu$$ occurs infinitely many times, being $S$ a set $\neq \emptyset$, and $\mu$ a measure (if you want to write this paragraph more rigurously please add your comments). Additionally the integrand is taken as different of a constant.

Example 1. See if you want my post in this MSE Bounding or evaluating an integral limit, as a first example of this kind of integrals.

Example 2. This afternoon I am trying to create a different example, with the help of Wolfram Alpha, from the sequence that starts as $$\int_0^1\int_0^1\frac{dxdy}{1-x\cdot\frac{1}{y}}=\int_0^1\int_0^1\frac{y}{y-x}dxdy,$$ $$\int_0^1\int_0^1\int_0^1\frac{z}{z-xy}dxdydz,$$ $$\ldots$$

I don't know if such examples that I've evoked were in the literature or are interestings. I presume that the importance of this kind of integrals should be to provide examples or counterexamples in some integration theory, or well show integrals with a nice closed-form, but I'm not sure because I've never seen this kind of integrals.

Question 1. Do you know if integrals as I've written in the first paragraph were in the literature as examples or counterexamples? Please if you know some source refers it, and I try to find the reference and read it. If you can add what should be the importance of this kind of integrals add your arguments. Many thanks.

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Question 2. Do you know or can you tell me tricks or methods to create more original examples? You can take as background integration in an abstract way as was written in the first paragraph. Many thanks.

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    You might want to look up for "functional integration" for examples where integration over $[0;1]^\infty$ arise : https://en.wikipedia.org/wiki/Functional_integration – Renart Sep 14 '17 at 14:52
  • Wow, many thanks tomorrow in the morning I am going to read your Wikipedia @Renart , and feel free if you want to edit your comment as an answer. –  Sep 14 '17 at 17:34

4 Answers4

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Integrating over an infinite number of variables means integrating over a measure space that is also an infinite-dimensional vector space. This operation is generically ill-defined:

There exist no infinite dimensional Lebesgue measure.

This does not mean that infinite-dimensional measures do not exist. All the examples I have encountered so far are related to Gaussian measures.

Concrete examples and applications.

  1. An infinite dimensional integral appears in the logarithmic Sobolev inequality of Leonard Gross, (see Adams and Clarke for a simplified proof). This inequality and its generalizations seem to be quite "hot" in the world of mathematical research, and according to Terry Tao's blog (see especially Exercise 10) they even appear in the extra-famous work of Perelman on the Ricci flow.
  2. Infinite dimensional integrals, in the form of the so-called Gibbs measures, appear in the statistical description of the flow of certain nonlinear equations: see Gigliola Staffilani's lecture notes, especially section 1.5.3.
Giuseppe Negro
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    True, Lebesgue measure does not exist on $\mathbb R \times \mathbb R \times \cdots$. But it does exist on $[0,1] \times [0,1] \times \cdots$. – GEdgar Sep 19 '17 at 15:03
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    @GEdgar: Interesting. Thank you for your answer and for this comment. I knew both concepts before (product measures and nonexistence of infinite-dim. Lebesgue measure), but I had never made the connection between them. – Giuseppe Negro Sep 19 '17 at 15:30
  • Many thanks, yours and the other users is a very nice work for all this community. –  Sep 19 '17 at 19:23
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If we define $$ f(n) := \underbrace{\int_0^1 ... \int_0^1}_{n} \frac{dx_1 \cdot ... \cdot dx_n}{1- x_1 \cdot ... \cdot x_n} $$ we then have that $f(2) = \frac{\pi^2}{6}$ and $\lim_{n\to \infty} f(n) = 1$.

Actually it is well known that $f(n) = \zeta(n)$ !

Zubzub
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  • Many thanks also to you, for your nice example. All answers were very nice and useful. –  Sep 19 '17 at 19:28
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There is a reasonable construction for a "product measure" where there are infinitely many factors, each a probability measure. In fact, in probability theory, when you talk about a sequence of random variables this construction is quite natural.

GEdgar
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An example could be perhaps constructed geometrically.

The integral $$ \int_{\mathcal{B}^{n}_1}\underbrace{ \mathrm{d}x_1 \mathrm{d}x_2 \cdots \mathrm{d}x_n}_{n \,\, \it{times}}=\frac{\pi^{n/2}}{\Gamma(\frac{n}{2}+1)}$$ where $\mathcal{B}^{n}_1$ stands for the $n$-ball of unitary radius, represents the volume of such $n$-ball.

Knowing that the volume of a $n$-ball approaches $0$ as $n$ tends to infinity, there might be a sense in which the curious "infinitely iterated" integral

$$ \int_{\mathcal{B}^{\infty}_1} \mathrm{d}x_1 \mathrm{d}x_2 \cdots =\lim_{n \to \infty} \int_{\mathcal{B}^{n}_1}\underbrace{ \mathrm{d}x_1 \mathrm{d}x_2 \cdots \mathrm{d}x_n}_{n \,\, \it{times}}= 0$$

could be defined.

An aedonist
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  • Many thanks for your attention and example. –  Sep 11 '17 at 14:46
  • As you see the integrand in your example is constant, but many thanks for your nice contribution. I hope that more users want to do contributions and we see those. Feel free if you want to add more notes. Good week. –  Sep 11 '17 at 15:06
  • I apologise, the detail about the integrand requested to be different from a constant completely escaled my attention. I will come up with an example complying to the requested characteristics. – An aedonist Sep 11 '17 at 15:08
  • Don't worry please, your example is very very nice. But add more examples or notes if you want. –  Sep 11 '17 at 15:12