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Which of

$\lim_{a\to\infty} \int_{-a}^{a}$ and $\lim_{r\to-\infty} \lim_{s \to \infty} \int_{r}^{s}$

does $\int_{\mathbb{R}}$ mean?

user27182
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  • See http://en.wikipedia.org/wiki/Improper_integral#Convergence_of_the_integral – lhf Mar 21 '13 at 17:47
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    Specifically, the "types of integrals" section, which details how an integral which is "improper" in the Reimann-integral sense can be proper as a Lebesgue integral. – Thomas Andrews Mar 21 '13 at 17:49
  • In some cases, $\int_{\Bbb{R}}$ stands for weaker notions of integration such as symmetric improper integral. Even in some cases, this notation is just exploited to denote a 1-dimensional distribution applied to a particular function. – Sangchul Lee Mar 21 '13 at 18:06
  • The first expression is called the Cauchy principal value of the integral and is usually denoted by $PV\ \int_{-\infty}^{+\infty} f(x), dx$ or $v.p.\ \int_{-\infty}^{+\infty} f(x), dx$. – Yury Mar 21 '13 at 18:32

2 Answers2

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Rather the second one: $$\int\limits_\mathbb{R}{f(x)\,dx}=\int\limits_{-\infty}^{\infty}{f(x)\,dx}=\lim_{r\to-\infty} \lim_{s \to \infty} \int_{r}^{s}{f(x)\,dx}$$

M. Strochyk
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It really depends on the context.

It is possible that it is just the Lebesgue integral, which is a "proper integral" without needing any limits.

As lhf said in comments, see the wikipedia article for improper integrals. Specifically, check out the section titled Types of Integrals.

If the function is non-negative and "nice" in some way, the various improper integrals and the Lebesgue integral will all agree.

The first form is the least likely. It has the obvious disadvantage that it potentially gives different results when $g(x)=f(x+c)$ for some constant $c$. It would also compute an integral of $0$ for any odd function. Would you want:

$$\int_{\mathbb R} xdx = 0?$$ That seems like an unfortunate result, somehow.

Possibly better than your double limit is to pick a fixed $a\in \mathbb R$ and define it as:

$$\lim_{s\to -\infty} \int_s^{a} + \lim_{r\to +\infty} \int_a^r$$

This has the advantage that it is symmetric in $r,s$ while the $\lim_r\lim_s$ definition is not (obviously) so. (The two limits are the same thing, it's just a little clearer that the $r,s$ limits are independent if you break at some $a$ as I have.)

Thomas Andrews
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