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I asked this question on Mathoverflow and received 6 downvotes so far and no answer.

In the comment section of this question several people seemingly freely assume that fundamental theorem of calculus should still hold for divergent integrals with improper bounds.

That is, they assume

$$\int_a^b f(\varphi(x))\varphi'(x) dx = \int_{\varphi(a)}^{\varphi(b)} f(u)du$$

holds even if $a$ or $b$ are $\pm\infty$ and the integral in the left-hand side is divergent to infinity.

I wonder, what can justify this assumption. In my view it is totally unjustified. Why do I think so? Because it allows the following:

$$I=\int_0^\infty1dx=2\int_0^\infty1du=2I$$ or (with substitution $u=2x$) $$\int_1^{+\infty}\frac1x dx=\int_2^{+\infty}\frac1u du$$

In the second case the indegrals even have different regularized values, the left-hand side integral has the regularized value $0$ while the right-hand side integral has the regularized value $-\ln2$. In other words, they are different divergent integrals with different properties.

In the comment section the user Johannes Hahn justified the relation with words "because it's true" but I do not see in what sense this equality even can be true for divergent integrals?

  • Is it equality of the values? If so, what value can have a divergent integral without extension of real numbers? Or an extension is assumed?

  • Is it equality of some other set of properties? In that case we see that the regularized values of these two integrals before and after substitution are different.

  • Something else?

Anixx
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    This is the change of variables formula, not the fundamental theorem for calculus. The fundamental theorem is $$\int_a^bf'(t)\text{d}t=f(b)-f(a).$$This may explain the downvotes too. Moreover, this kind of stuff is way too easy for MO, this belongs here, so this is another reason for the downvotes. – Anthony Nov 03 '20 at 19:37
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    $$\int_0^\infty1dx=2\int_0^\infty1du$$ can be rewritten $+\infty = +\infty$, and $$\int_1^{+\infty}\frac1x dx=\int_2^{+\infty}\frac1u du$$ can be rewritten $+\infty = +\infty$. I see no contradiction here. – TheSilverDoe Nov 03 '20 at 19:40
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    It certainly does not make much sense to manipulate diverging integrals. So, this question has some point although it is about a substitution-rule and not about the fundamental theorem of calculus. – Peter Nov 03 '20 at 19:43
  • @AnthonySaint-Criq the fundamental theorem in its basic form assumes continuous function on the closed interval [a,b] https://www.sfu.ca/math-coursenotes/Math%20158%20Course%20Notes/sec_ImproperIntegrals.html I also do not see how this equality can hold without extension of real numbers given improper integrals have no real value. – Anixx Nov 03 '20 at 19:44
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    Here is a formal statement, which I didn't mean to give in a comment. But still, this doesn't do anything with the fact that you're talking about the change of variables, not this theorem. Change of variables also asserts that the nature of the two integrals is the same (both convergent or divergent), so there's no issue whatsoever with your question. – Anthony Nov 03 '20 at 19:46
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    @AnthonySaint-Criq The second reason you mention (about the level) might be a reason , but a very bad one. Anyway, this site tends to unjustified negative feedback, which I know from many questions I posted there. – Peter Nov 03 '20 at 19:48
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    @Peter Of course I disagree with how easy it is to get downvoted for this reason. But facts are such questions get downvoted regardless of the quality because of them being too easy for MO. People do this over there, and again, I disagree with the method but here is how it works ! – Anthony Nov 03 '20 at 19:50
  • @AnthonySaint-Criq how that formal statement can be even applied to the values of divergent integrals, if they do not have a real value? Or you assume an extended real line? The linked theorems do not! – Anixx Nov 03 '20 at 19:52
  • @Anixx the theorem (fundamental theorem) is not valid for $a=-\infty$ or $b=+\infty$ and the function is required to be regular at $a$ and at $b$. (these hypothesis I didn't want to put in my first comment because it wasn't the point). However, the change of variables formula holds in this more general setup. I suggest you have a look at this for an example ! – Anthony Nov 03 '20 at 19:54
  • @AnthonySaint-Criq I repeat my question, how the change-of-variable formula can hold (or even what does it mean) for divergent integrals if they do not take values from real numbers? If some extension of real numbers is assumed, like this https://en.wikipedia.org/wiki/Extended_real_number_line then the formula is irrelevant because I did not assume this extension in my original linked question. – Anixx Nov 03 '20 at 19:57
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    I have already mentioned my opinion about maniplulations of diverging integrals. We transform a divergent integral in another diverging integral , whether mathematically correct or not, I would avoid this. – Peter Nov 03 '20 at 19:59
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    But you didn't specify which setting you use. Using the extended real numbers is the standard, and if you use a different setting, you need to explicitly specify that setting. – Daniel Fischer Nov 03 '20 at 20:00
  • Because $\displaystyle\int_a^bf(t)\text{d}t$ converges if and only if $\displaystyle\int_{\varphi^{-1}(a)}^{\varphi^{-1}(b)}f\circ\varphi(t)\varphi'(t)\text{d}t$ converges. For positive functions at least, you are just writing, in the case of divergent integrals, $+\infty=+\infty$, so this still makes sense. But for functions with non-constant sign, or for complex-valued functions, there's nothing more than the nature of the integrals to tell something about – Anthony Nov 03 '20 at 20:02
  • @DanielFischer "Using the extended real numbers is the standard" - since whence? When I studied in the university we never used it. Also, as you can see, the linked question explicitly asks in the last phrase "So, I wonder, if there is any intuition or example from other fields where these integrals demonstrate equivalence in a more strict sense than being both divergent to infinity?" – Anixx Nov 03 '20 at 20:03
  • @Anthony Saint-Criq my last sentence in the question reads, "So, I wonder, if there is any intuition or example from other fields where these integrals demonstrate equivalence in a more strict sense than being both divergent to infinity?", so I was NOT asking whether they are convergent or not. – Anixx Nov 03 '20 at 20:05
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    But an integral need not be "divergent to infinity". Integrate $\cos$ over the real line for instance. To me, your question now looks like a $1+2+3+...=-1/12$ kind of thing... – Anthony Nov 03 '20 at 20:10
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    I guess the crucial term here is that the original MO post talks about the integrals being "regularized" to a certain value - and it is certainly not clear to me what it means. –  Nov 03 '20 at 20:10
  • @StinkingBishop some people say, this question is too easy?.. Integrals can be regularized, like series can. – Anixx Nov 03 '20 at 20:13
  • Then just say what you mean by regularization ? Cesaro ? Cauchy principal value ? Anything else ? Also, try to edit your question title/body to take into account the comment section – Anthony Nov 03 '20 at 20:14
  • @AnthonySaint-Criq one of the ways is to convert the integral into a series and find its regularized value. I use this Mathematica code for an integral of f(x) from 0 to infinity: Sum[f[s x],{x,1,Infinity},Regularization->"Borel"]//FullSimplify

    Limit[s %,s→0]

     – Anixx Nov 03 '20 at 20:18
  • @AnthonySaint-Criq one also can use Fourier transforms https://math.stackexchange.com/questions/3500680/when-the-following-property-of-integrals-holds (natural integral at zero with the opposite sign, for holomorphic functions) In general, the question of the best formula is open: https://mathoverflow.net/questions/369378/more-or-less-universal-formula-for-regularization-of-divergent-integrals – Anixx Nov 03 '20 at 20:22
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    I don't know since when. I'd guess, about the time when Lebesgue theory became standard, since the extended real numbers are a most convenient setting for that. Of course there's also the standard that says $\int_{-\infty}^{+\infty} 1,dx$ doesn't exist rather than that it's $+\infty$. But in my experience, that is mainly used before Lebesgue theory is treated. – Daniel Fischer Nov 03 '20 at 20:25
  • @TheSilverDoe $+\infty$ and $-\infty$ are not numbers unless you extend the set of real numbers. – Anixx Nov 03 '20 at 20:42
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    @Anixx If you don't allow integrals to be equal to $+\infty$, then $\int_0^\infty1dx$ and $\int_1^{+\infty}\frac1x dx$ just have no sense. But you wrote them, so I assumed you gave them the only sense you can give them, i.e. to attribute them the value $+\infty$. – TheSilverDoe Nov 03 '20 at 20:43
  • @TheSilverDoe the integrals are divergent integrals, so the have sense in multiple ways (regularized value, rate of growth, germ at infinity and others). Also read the linked question. – Anixx Nov 03 '20 at 20:46
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    @Anixx Which sense gives you to $\int_0^\infty1dx$, if you don't give it the value $+\infty$ ? – TheSilverDoe Nov 03 '20 at 20:48
  • @TheSilverDoe it is a divergent integral, equal to $\int_0^\infty 1/x^2 dx$. See the linked question, again. – Anixx Nov 03 '20 at 20:50
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    @Anixx So it is not a number, nor $+\infty$ ?? And how do you define $\int_0^1 1/x dx$ ?? Please give a precise definition of the symbol $\int$ for functions that are not integrable, without using the symbol $\int$ itself in the definition. – TheSilverDoe Nov 03 '20 at 20:52
  • @TheSilverDoe I define then as divergent integrals, lol. What other definition is needed? Yes, one can call them numbers, but not real numbers, of course. – Anixx Nov 03 '20 at 20:53
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    @Anixx Ok, so you are not able to give me the definition of the object about which you are asking. There is no point in argumenting further. I can also ask a question : "why is cdsjcnds equal to nfjknvfjksnvk and not to ,djzdz ??", and when one asks "what is cdsjcnds ?", answer "I define it as cdsjcnds, lol." – TheSilverDoe Nov 03 '20 at 20:54
  • @TheSilverDoe the definition is simple: it is an integral of a function that diverges to infinity (has limit infinity or if you want, a sequence of consequtive Riemannian or Lebesgue sums that infinitely grows). – Anixx Nov 03 '20 at 20:56
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    @Anixx Are you aware that you use the word "integral" to define what an integral is ?? – TheSilverDoe Nov 03 '20 at 20:57
  • @TheSilverDoe no, I use the word "sequence" to define what integral is, but since it is already defined elsewhere, I think I can use the term "integral" without defining it from scratch. – Anixx Nov 03 '20 at 20:59
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    @Anixx So it is the limit of a sequence that tends to $\infty$, but it is not $\infty$. Interesting. – TheSilverDoe Nov 03 '20 at 21:01
  • @TheSilverDoe I did not ever say "limit". One can think about it as a limit, but it would be another, generalized definition of limit. A sequence itself specifies a divergent integral, and those sequences that approach each other (difference between them vanishes) are equal. Similar to how a convergent sequence defines a real number. – Anixx Nov 03 '20 at 21:04
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    @Anixx So you define an integral as an equivalence class of sequences of Riemann sums ? Ok, then, go ahead. I will not follow you there. – TheSilverDoe Nov 03 '20 at 21:12
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    I already flagged this question, it is still pending. At this point, seeing the exchanges between you two, I'm getting the feeling that this question is "unlikely to be salvaged by any means", more than just "in need for clarification"... In this current state, this question just deserves to be closed. @Anixx, please try to clarify clearly and precisely what you mean, and also correct the mistakes you made here and there. It took 36 comments to roughly have you say that you don't know how you define things, which is up to no good. – Anthony Nov 03 '20 at 22:18
  • @AnthonySaint-Criq I already got the answer, so I can post my own answer. This question is not needed any more. – Anixx Nov 03 '20 at 22:19

3 Answers3

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If I change variable/u-substitute in a divergent improper integral, I get another divergent improper integral. Is that not intuitively obvious? A change of variable shouldn’t be able to turn divergent improper integrals into convergent ones, or vice versa.

Changes of variable should also preserve the manner of divergence -- whether the original improper integral approaches $+\infty$, approaches $-\infty$, or oscillates forever (or some combination), the $u$-substituted version is going to be doing the exact same thing.

Edit: Let's take the example from the original post, where we change variable in $\int_1^\infty \frac{1}{x} dx$ with $u = 2x$. We're supposed to get $du = 2x dx$, $du/2 = dx$, and $$\int_1^\infty \frac{1}{x} dx = \int_2^\infty \frac{1}{u/2} \frac{du}{2} = \int_2^\infty \frac{1}{u} du.$$

It seems that OP doesn't like when people say, informally, "The value of the improper integral here is $+ \infty$." In fact, that's widely understood to mean a certain kind of limiting behavior which gives more specificity over just saying, "The value of the improper integral does not exist", or, "The improper integral does not converge to a value." Let's go back to high school calculus and remember how these improper integrals are computed: $$"\int_1^\infty \frac{1}{x} dx" \text{ means/equals/is defined as } "\lim_{b \to \infty} \int_1^b \frac{1}{x} dx, \text{ if it exists."}$$

To say that the improper integral "is" $+ \infty$, is just to say that that limit "is" $+ \infty$, right? That is, as $b \to \infty$, the value of the integral is increasing without bound. (I'm putting "is" in quotes here to remind us that the $+ \infty$ is really describing a limiting behavior and not a numerical value.)

So let's look at what that means for our $u$-substituted integral:

$$\int_1^\infty \frac{1}{x} dx = \lim_{b \to \infty} \int_1^b \frac{1}{x} dx = \lim_{b \to \infty} \int_2^{2b} \frac{1}{u} du = \int_2^\infty \frac{1}{u} du,$$

and clearly if one of those limits "is" $+ \infty$, then the other limit "is" $+ \infty$ too. The same would hold true regardless of which specific $u$-substitution I made in the original--the corresponding limit is going off to $+ \infty$.

Note that I haven't said anything about the "regularized values" of these integrals. The "regularized value" is not to be confused with the integral itself. In fact, if the integral diverges but it has a "regularized value", then the value is already, by definition, different from the integral itself! There's no reason a change of variable should preserve whatever "regularized" number is assigned to a divergent integral, much less to think that this casts fundamental doubt on $u$-substitution itself.

Rivers McForge
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  • Yes, this is obvious. Still, the properties of those integrals may be different. – Anixx Dec 24 '20 at 01:32
  • @anixx Ok, and if one approaches $+\infty$, then the other one should also, correct? Changing variable doesn’t change the value of the integral. If it does, then you’re doing it wrong. – Rivers McForge Dec 24 '20 at 01:36
  • How can you speak about the "value" of an improper integral without extending real numbers? There is no real number that is equal to an improper integral, so about what value do you speak? The regularized value changes, as shown in the example. – Anixx Dec 24 '20 at 01:42
  • If you assume the extended real line ${\overline {\mathbb {R} }}$ https://en.wikipedia.org/wiki/Extended_real_number_line then yes, the value of all divergent to infinity integrals is the same. But you should explicitly declare this real line extension (there can be other variants). – Anixx Dec 24 '20 at 01:45
  • “There is no real number that is equal to an improper integral” is false, unless you’re solely talking about divergent improper integrals. If an improper integral diverges to $+ \infty$, then so does any u-substituted equivalent. Don’t confuse the “regularized value of the integral” with the integral per se—two improper integrals that both diverge to $+ \infty$ might have very different “regularized values” depending on how the regularization is being carried out. – Rivers McForge Dec 24 '20 at 01:49
  • Yes, I meant divergent integrals, sure. – Anixx Dec 24 '20 at 01:51
  • There is usually the proper regularized value (the same between the major methods). – Anixx Dec 24 '20 at 01:52
  • But the regularized value is different from the integral itself. Correct? There’s no reason a change of variable should preserve regularized values, because what changes of variable do is they preserve the original integral value. If the original improper integral becomes unboundedly positive (“diverges to $+ \infty$“), then the u-substituted one will also. – Rivers McForge Dec 24 '20 at 01:56
  • What is "original integral value" of a divergent integral? Do you assume $\infty$ or $-\infty$ are values? On what basis? – Anixx Dec 24 '20 at 01:59
  • @anixx Let me flesh out what I’m saying in my answer per se. – Rivers McForge Dec 24 '20 at 02:00
  • I think this is a perfect answer – Riemann'sPointyNose Dec 24 '20 at 02:36
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Taken from MathOverflow's help section:

MathOverflow's primary goal is for users to ask and answer mathematical questions related to current research in mathematics.

So the downvotes are definitely warranted as this is not related to current research in mathematics.

Now let's address the crux of your argument. First note that improper integrals are limits and $$\int_0^\infty\mathrm dx = \lim_{b\to\infty}\int_0^b\mathrm dx=\infty$$ Which means that the limit does not exist and the improper integral diverges. So what does it mean when an integral or a sum diverges? It basically means that neither of the two classical summability methods worked. Now if a different summability method is applied, then it must be explicitly stated. Note that different summability methods can report different values for divergent integrals or sums. To my surprise, all of this was already explained to you here. Let me know if you have any other questions.

k170
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  • It is related to current research in mathematics. – Anixx Dec 24 '20 at 00:58
  • I think, everything is already explained in the question. – Anixx Dec 24 '20 at 01:01
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    No. Understanding the basics of limits and u-substitution is not related to current research in mathematics. – k170 Dec 24 '20 at 09:53
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    Technically speaking, literally every mathematical statement is, or could be claimed to be, "related to" current research in mathematics. Doesn't mean MathOverflow is the right platform for it. – Rivers McForge Dec 24 '20 at 18:48
  • @RiversMcForge, my point exactly. This question isn't asking about some cutting edge theorem that's pushing mathematical understanding to a new level. This is a question about learning the basics of mathematics, which is fine on this platform. – k170 Dec 24 '20 at 23:50
  • @k170 it is related to this research: https://mathoverflow.net/questions/115743/an-algebra-of-integrals/342651#342651 – Anixx Dec 27 '20 at 15:47
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I think @Rivers McForge gave you a perfect answer to this, but I thought I'd try to maybe rephrase it a different way.

Ok the whole answer has been edited.

The question you asked (which I think is a completely valid and good question) is why do we say $$ \int_{a}^{b}f(\phi(x))\phi'(x)dx = \int_{\phi(a)}^{\phi(b)}f(u)du $$ if the integral diverges? Don't you end up with nonsense? The answer is yes - you do end up with nonsense. We can only really technically use an equals sign on numbers (or similar set of objects), so if neither integral converges, then using an equals sign is meaningless.

The point to make is, that if the left hand side is nonsense, the right hand side will also be nonsense, and vice versa. In this sense, it is consistent (by consistent what I mean is you don't end up with number = nonsense, you can only ever get number = number or nonsense = nonsense). So people continue (and are comfortable with) using the equals sign still.

What you did with your two examples is manipulate expressions that are essentially "nonsense = nonsense", which doesn't make sense. You can only algebraically manipulate or do things with expressions that are well-defined quantities, which they were not to begin with. The point is that if we know one side of the formula is defined, so is the other. And that is all. Nothing more, nothing less.

Last thing to mention - regularised values for divergent sums and integrals play by a completely different set of rules, so please don't use them for arguments like this. There's no reason to think that substitution should preserve regularised values.

Riemann'sPointyNose
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