Integral from infinity to infinity

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Solution 1

This is not necessarily true. Take the following example; $$\int_a^{2a}\frac1x\mathrm{d}x=[\ln{|x|}]_a^{2a}=\ln{(2)}$$ If we take $a\to\infty$ then the integral becomes $$\int_\infty^\infty\frac1x\mathrm{d}x=\ln{(2)}$$ as the integral is constant for all $a\in\mathbb{R}$. What I guess your professor meant was that $$\lim_{a\to\infty}\int_a^a f(x)\mathrm{d}x=0$$ which is trivially true as the LHS is constantly zero.

Solution 2

An improper integral with an endpoint of $\infty$ means a limit of proper integrals where the endpoint approaches $\infty$. Thus a reasonable definition of $\int_{\infty}^\infty f(x)\; dx$ would be $$ \int_{\infty}^\infty f(x)\; dx = \lim_{a, b \to \infty} \int_a^b f(x)\; dx $$ This is $0$ if and only if $\int_a^\infty f(x)\; dx$ converges for some $a$.

EDIT: If the double limit is $0$, there is $N$ such that $\left|\int_a^b f(x)\; dx\right| < 1$ for all $N < a < b$. For any $\epsilon > 0$ there is $M > N$ such that for $b, c > M$, $$ \left|\int_b^c f(x)\; dx \right| = \left| \int_a^c f(x)\; dx - \int_a^b f(x)\; dx \right|< \epsilon$$
and this implies that $\lim_{b \to \infty} \int_a^b f(x)\; dx$ exists, i.e. $\int_a^\infty f(x)\; dx$ converges.

Conversely, if $\int_a^\infty f(x)\; dx = L$ converges, then for any $\epsilon > 0$ there is $N$ such that $\left|\int_a^b f(x)\; dx - L\right| < \epsilon/2$ whenever $b > N$. Then if $b > N$ and $c > N$, $$ \left| \int_b^c f(x)\; dx\right| = \left|\int_a^c f(x)\; dx - \int_a^b f(x)\; dx \right| < \epsilon $$

Solution 3

As has been pointed out by other answers, this is not always true because the symbol $\infty$ can hide many things, even if we work with the extended reals. The actual meaning of the $\infty$ is a limiting process as a certain variable becomes arbitrarily large. The upper and lower limits in the $$\int_a^b f(x)\mathrm d x$$ however may approach $\infty$ at different rates, and this is the key point. Your professor's argument works only when $a$ and $b$ approach infinity equally fast, or in other words, when they are of equal order at infinity. In particular it is always valid if $a=b.$

So what about your professor's "proof"? Well, the ambiguity should now be obvious -- he uses the same symbol $\infty$ for things that may behave differently. I think it's safe to assume he's thinking only of the case when the variables are of equal order at infinity. Otherwise his proof breaks down since $\infty-\infty$ can then be anything.

PS. However, you say an integral from $\infty$ to $\infty$ has no meaning to you. Well, I see you're thinking of the usual ordering of the reals here. But note that we're not just dealing with the reals here, but the extended reals. As explained above, the best way to think of it is to think of the limits of the integral as approaching infinity (at not necessarily equal rates). Then it's easy to make sense of it. Another way may be to think of the one-point compactification of the real axis.

Solution 4

I disagree with the answers here. From Lebesgue's perspective, we can think of the integral under question as an integral over the set of all real $x$ satisfying $\infty < x < \infty$. That's an integral over the empty set, which is always 0.

Solution 5

As Peter Foreman mentioned, there are some occasions where you will get an integral of this form, however the exact integral does matter. Firstly visualise the error function: $$\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{-t^2}dt$$ notice that: $$\lim_{x\to 0}\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^0e^{-t^2}dt$$ Normally we can visualise an integral as area, but in this form it does not make sense since the range over which the area found $\to0$. The most obvious case when this integral is zero, no matter the function, would be when both the bounds are equal as the range of the integral is then zero. Overall, it comes down to notation and making sure the bounds are properly defined.

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Updated on August 23, 2022

Comments

  • Victor
    Victor over 1 year

    My physics professor today wrote on the blackboard: $$ \int_{\infty}^{\infty} f(x) dx = 0 $$ for every function $f$. And the proof he gave was: $$ \int_{\infty}^{\infty} f(x) dx = \int_{\infty}^{a} f(x) dx + \int_{a}^{\infty} f(x)dx = - \int_{a}^{\infty} f(x) dx + \int_{a}^{\infty}f(x)dx = 0$$

    However I'm still not convinced, for me an integral from infinity to infinity has no meaning. Therefore, what I'm asking is: does the above equations make sense? If not, are there cases where they do make sense? I'm thinking about functions that converge to 0 in $+\infty$.

    EDIT: Actually, the function f considered was a density, i.e.: $$ \int_{-\infty}^{+\infty} f(x)dx = 1 $$ and $f(x) \geq 0$ for all $x$.

    • José Carlos Santos
      José Carlos Santos over 4 years
      Not much IMHO. ${}$
    • Ethan Bolker
      Ethan Bolker over 4 years
      I hope your professor meant this as a joke. Did he (or she) actually go on to use this result in an argument?
    • copper.hat
      copper.hat over 4 years
      How can the integral be 0 and 1 at the same time?
    • Victor
      Victor over 4 years
      @copper.hat one integral is from -infinity to infinity, the other one is from +infinity to +infinity
    • copper.hat
      copper.hat over 4 years
      I see. It is a bit meaningless. I would not spend too much time pondering it. It is like $\int_a^a$ where $a$ is finite.
    • Greg Martin
      Greg Martin over 4 years
      The main point is that $\int_\infty^\infty$ never appears in practice in mathematics. (If it were going to be a way of abbreviating some naturally occurring expression, then use that expression itself instead.) Thinking about logical consequences of definitions can be useful and fun for its own sake, but let's not mistake it for doing mathematics.
    • GEdgar
      GEdgar over 4 years
      Confused students sometimes do get an integral $\int_\infty^\infty$ when they make a substitution incorrectly.
    • Zacky
      Zacky over 4 years
      @Asaf Karagila can you please explain why have you taken down this question from HNQ?
    • Asaf Karagila
      Asaf Karagila over 4 years
      @Zacky: I used to put more effort into making the titles of questions better. In 99% of the time it would include putting MathJax, which would make it HNQ ineligible anyway. Admittedly, I had gotten lazy. But also a bit discontent when some users reversed MathJax titles for the sole reason that they wanted their questions back on the HNQ. The title of this question is awful, it is not descriptive, and it can be greatly improved by making it more explicit, in that process, MathJax would have to be added anyway.
  • Victor
    Victor over 4 years
    could you provide more details on the condition $\int_a^{\infty} f(x)dx$ converges for some $a$ is a IFF condition
  • Peter Foreman
    Peter Foreman over 4 years
    How would you define $a,b\to\infty$ because as I have shown, a suitably chosen $b(=2a)$ causes the integral to remain non-zero.
  • Robert Israel
    Robert Israel over 4 years
    @PeterForeman In the usual way: the limit is $L$ if for every $\epsilon > 0$ there is $N$ such that for all $a>N$ and $b > N$, $\left| \int_a^b f(x)\; dx - L \right| < \epsilon$. In your example the limit does not exist.
  • Allawonder
    Allawonder over 4 years
    I thought I read somewhere that Lebesgue's integral is an extension of Riemann's in such a way that whenever the latter exists, the former agrees with it. From Riemann's viewpoint, for example, evaluating $$\int_{\infty}^{\infty}{\frac 1 x} \mathrm d x$$ as shown in one of the answers above, gives a nonzero real number, namely $\log 2.$ Yet here you claim that Lebesgue's method gives $0.$ If what I remember reading is actually the case, then how do you explain this discrepancy? Well, I think the main problem is in the fact that the symbol $\infty$ can mean many things, depending on context.
  • Allawonder
    Allawonder over 4 years
    How do you explain the inequality $\infty<x,$ for example, from the set-theoretic point of view? I understand the other inequality, namely $x<\infty,$ is usually used to mean that $x$ is a finite real number. But note that this infinity is not signed. That is, $x$ can be an arbitrary (finite) real number. Thus $\infty$ here stands for what's usually written as $\pm \infty.$ So how would you explain the other inequality? I don't think $x$ is not a finite real number would do, since that would logically be written $\infty \le x.$ So, you see, there are many issues hidden here.
  • Theo Bendit
    Theo Bendit over 4 years
    You shouldn't be applying the FTC to an unbounded, discontinuous function like that.
  • goblin GONE
    goblin GONE over 4 years
    @Allawonder, in my opinion, there are no issues hidden here, at least from my point of view (I have roughly a masters degree in pure mathematics). In particular, we begin by defining the affinely extended real line $[-\infty,\infty]$ by taking $\mathbb{R}$ as adjoining two points, namely $\infty$ and $-\infty$. We then extended our functions and operations on $\mathbb{R}$ to make them functions on $[-\infty,\infty]$. The inequality $\infty < x$ can be interpreted as follows. Firstly, $\infty$ is one of the elements of $[-\infty,\infty]$. Secondly...
  • goblin GONE
    goblin GONE over 4 years
    Secondly, $<$ is the version of $<$ that we obtained by taking the usual $<$ operation on $\mathbb{R}$ and extending it to $[-\infty,\infty]$ in the usual way. Thirdly, $x$ is a real number, and hence an element of $[-\infty,\infty]$. Thus what we find is that $\infty < x$ is simply false. Hence the set of all such $x$ is the empty set.
  • Allawonder
    Allawonder over 4 years
    The inequality depends on how you define $\infty$ and extend the usual operations to it. If we extend the real line to $[-\infty,+\infty]$ and then identify the endpoints, and denote them by $\infty,$ then it is in fact true that $\infty\lt x$ for any real $x.$ But that's not the main issue here. The first thing I noticed is that your disagreement has nothing to do with how the integral is defined (Riemann or Lebesgue). The point of disagreement arises in how we differently interpret the symbol $\infty.$ If you take it as an abbreviation for a limiting operation (which is what most of the...
  • Allawonder
    Allawonder over 4 years
    ...answers above do), then it isn't always $0$ for all integrands. But if you think of the real line extended by one point as I described above, then it's trivially true since $$\int_a^a f(x) \mathrm d x=0.$$ Also, this triviality, I think, is why most other answers took instead the other interpretation of the symbol $\infty,$ in which case the professor's argument is false. Thus, again, the problem has everything to do with the ambiguity of what the professor means by $\infty.$