7
$\begingroup$

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 $x$.

$\endgroup$
  • 4
    $\begingroup$ Not much IMHO. ${}$ $\endgroup$ – José Carlos Santos Jun 25 at 16:12
  • 4
    $\begingroup$ I hope your professor meant this as a joke. Did he (or she) actually go on to use this result in an argument? $\endgroup$ – Ethan Bolker Jun 25 at 16:21
  • $\begingroup$ How can the integral be 0 and 1 at the same time? $\endgroup$ – copper.hat Jun 25 at 16:36
  • 1
    $\begingroup$ @copper.hat one integral is from -infinity to infinity, the other one is from +infinity to +infinity $\endgroup$ – Victor Jun 25 at 16:37
  • 1
    $\begingroup$ @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. $\endgroup$ – Asaf Karagila Jul 13 at 12:22
22
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ You shouldn't be applying the FTC to an unbounded, discontinuous function like that. $\endgroup$ – Theo Bendit Jun 26 at 10:13
6
$\begingroup$

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 $c > N$, $$ \left| \int_b^c f(x)\; dx\right| = \left|\int_a^c f(x)\; dx - \int_a^b f(x)\; dx \right| < \epsilon $$

$\endgroup$
  • $\begingroup$ could you provide more details on the condition $\int_a^{\infty} f(x)dx$ converges for some $a$ is a IFF condition $\endgroup$ – Victor Jun 25 at 18:59
  • $\begingroup$ 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. $\endgroup$ – Peter Foreman Jun 25 at 19:37
  • $\begingroup$ @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. $\endgroup$ – Robert Israel Jun 25 at 21:46
3
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Allawonder Jun 26 at 8:37
  • $\begingroup$ 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. $\endgroup$ – Allawonder Jun 26 at 8:43
  • $\begingroup$ @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... $\endgroup$ – goblin Jun 26 at 11:46
  • $\begingroup$ 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. $\endgroup$ – goblin Jun 26 at 11:48
  • $\begingroup$ 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... $\endgroup$ – Allawonder Jun 27 at 17:57
0
$\begingroup$

I believe your professor is correct under the conditions that $\int_{-\infty}^{+\infty} f(x)dx = 1$ and $f(x) \geq 0$ for every $x\in\mathbb{R}$. I will flesh out his proof in greater detail below, but first I want to show how we can make sense of $\int_{\infty}^{\infty} f(x) dx$:

First, let $f$ be a density function such that $\int_{-\infty}^{+\infty} f(x)dx = 1$ and $f(x) \geq 0$.

Now, for every $a,b\in\mathbb{R}$ such that $a\lt b$, consider $\int_{a}^{b} f(x) dx=F(a)-F(b)$ where $F'(x)=f(x)$, by FTOC.

Observe $F(a)-F(b)=g(a,b)$. Then, by substitution we have $\int_{a}^{b} f(x) dx=g(a,b)$. Next, we take the limit of this equation as $b\to\infty$ so that

$lim_{b\to \infty}g(a,b)=lim_{b\to \infty}\int_{a}^{b} f(x) dx=\int_{a}^{\infty} f(x) dx$

by definition of an improper integral. We again take the limit of the equation, this time as $a\to\infty$, so that

$\lim_{a\to \infty}\{lim_{b\to \infty}g(a,b)\}=\lim_{a\to \infty}\{\lim_{b\to \infty}\int_{a}^{b} f(x) dx\}=\lim_{a\to \infty}\int_{a}^{\infty} f(x) dx=\int_{\infty}^{\infty} f(x) dx$

So really this odd looking integral $\int_{\infty}^{\infty} f(x) dx$ is simply saying that we are taking the nested limit of some bivariate function, $g(a,b)$, as $a\to\infty$ and $b\to\infty$. Now let's prove $\int_{\infty}^{\infty} f(x) dx=0$ as follows:

$\int_{\infty}^{\infty} f(x) dx = \lim_{a\to \infty}\{\lim_{b\to \infty}\int_{a}^{b} f(x) dx\}$

$=\lim_{a\to \infty}\{\lim_{b\to \infty}[\int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx]\}$ where $c$ is some real constant such that $a\lt c\lt b$, by properties of definite integrals

$=\lim_{a\to \infty}\{\int_{a}^{c} f(x) dx + \lim_{b\to \infty}\int_{c}^{b} f(x) dx\}$ by the constant function rule for limits

$=\lim_{a\to \infty}\int_{a}^{c} f(x) dx + \lim_{b\to \infty}\int_{c}^{b} f(x) dx$ by the constant function rule for limits

$=\lim_{a\to \infty}-\int_{c}^{a} f(x) dx + \lim_{b\to \infty}\int_{c}^{b} f(x) dx$ by properties of definite integrals

$=-\lim_{a\to \infty}\int_{c}^{a} f(x) dx + \lim_{b\to \infty}\int_{c}^{b} f(x) dx$ by the constant multiple rule for limits

$=-\lim_{a\to \infty}[F(a)-F(c)] + \lim_{b\to \infty}[F(b)-F(c)]$ by FTOC

$=-[\lim_{a\to \infty}F(a)-F(c)] +\lim_{b\to \infty}F(b)-F(c)$ by the constant function rule of limits

$=-\lim_{a\to \infty}F(a)+F(c) +\lim_{b\to \infty}F(b)-F(c)$ by the distributive law

$=\lim_{b\to \infty}F(b)-\lim_{a\to \infty}F(a) + F(c)-F(c)$ by the commutativity law

$=\lim_{b\to \infty}F(b)-\lim_{a\to \infty}F(a)$ by the additive inverse and identity laws

Since $f$ is a density function such that $\int_{-\infty}^{+\infty} f(x)dx = 1$ and $f(x) \geq 0$ for every $x\in\mathbb{R}$, then $\lim_{x\to \infty}F(x)$ exists, or in other words, $\lim_{x\to \infty}F(x)=L\in\mathbb{R}.$ Now, $a$ and $b$ are just arbitrary variables whose values increase positively without bound. Hence, $\lim_{b\to \infty}F(b)=\lim_{a\to \infty}F(a)=L$. By substitution, we have

$=L-L=0$

Therefore, if $\int_{-\infty}^{+\infty} f(x)dx = 1$ and $f(x) \geq 0$, then $\int_{\infty}^{\infty} f(x) dx =0$

(to confirm properties of definite integrals used in this proof, see http://mathworld.wolfram.com/DefiniteIntegral.html.)

$\endgroup$
0
$\begingroup$

It makes perfectly sense to integrate from -inf to inf - just look at a probability density, e.g. the normal distributions density.

$\endgroup$

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.