When discussing divergent integrals with people, I got curious about the following:

Is there an $\mathbb{R}$-algebra $A$ together with a map (could be defined on just a subspace)

$$\int_0^{\infty}: C^{\infty}(\mathbb{R})--\to A$$

which behaves like integration, and is defined even for some function whose integration is divergent in the usual sense? Or,

Can we find some universal target of integration which is like the module of Kahler differentials as the universal target of derivation?

In other words, is there an $\mathbb{R}$-algebra $A$ together with a map $T: C^{\infty}(\mathbb{R})\to A$ such that (here is a list of plausible properties)

$$T(f(x))=\int_0^{\infty}f(x)dx\in\mathbb{R},\text{ if the RHS converge}.$$ $$T(f(x))-T(f(x+a))=\int_0^af(x)dx \text{ for any }f(x).$$ $$T(f(x))=aT(f(ax)) \text{ for } a>0.$$

(Edit: I removed the 4th, which is included in the first.)

Or for some reasons such an $\mathbb{R}$-algebra could only be $\mathbb{R}$? I tried to construct a ``free'' algebra of some kind but it is not clear to me what I got. (from the conditions above there are too many generators and relations, and there are even things from "integration by parts" given the last rule, I'm not sure what the quotient of the generators by the relations gives.)

(Edit: people asked why $A$ need to be an algebra, I don't have a good reason for that. I just want to see if one can extend the definition of integration so that they land in some vector space with a certain structure. The most naive thing I can think about is that the result of an integral is a certain kind of "number", and we add, subtract, multiply numbers.)

  • $\begingroup$ Something like defining $\int_0^\infty e^{-cx} dx = 1/c$ even for $c=it$ with $t \in {\bf R}^*$? (Take real and imaginary parts if you insist on having an algebra of real- rather than complex-valued functions.) $\endgroup$ Commented Dec 8, 2012 at 5:47
  • $\begingroup$ I don't insist having something over $\mathbb{R}$. I just want to see if these kind of things may work. $\endgroup$
    – 36min
    Commented Dec 8, 2012 at 6:23
  • $\begingroup$ Your first and fourth conditions are equivalent. $\endgroup$
    – S. Carnahan
    Commented Dec 8, 2012 at 9:30
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    $\begingroup$ This is a bit of a nit-pick but perhaps also interesting: you should perhaps say what you mean by 'the usual sense' since there is a difference between the (improper) Riemann integeral and the Lebesgue integral over (0,infty) being finite. If memory serves well t^(-1) sin(t) is an example. $\endgroup$
    – user9072
    Commented Dec 8, 2012 at 13:51
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    $\begingroup$ Maybe it would be easier to start by doing this for (possibly divergent) series. Or, equivalently, sequences. $\endgroup$ Commented Dec 8, 2012 at 14:32

6 Answers 6


Property (2) gives $T(1) = T(1)+a$ for any real $a$, which is not solvable in any real algebra (or vector space) $A$. Property (3) leads to a similar issue as it implies that $T(1) = aT(1)$ for all $a>0$.

Note that many common ways of evaluating divergent sums and integrals (e.g. zeta function regularisation) do not actually obey (2) or (3). For instance, the famous identity $1 + 2 + 3 + \ldots = -1/12$, which is valid if the LHS is summed using zeta function regularisation, is inconsistent with basic axioms such as (2), as discussed in this blog post of mine. Also, none of these methods are able to sum all divergent series (or integrate all non absolutely integrable functions). In view of this, I doubt that an axiomatic approach that assumes that all integrals can be integrated is the most natural way to proceed here.

ADDED LATER: Using enough abstract nonsense, one can integrate arbitrary functions, but in a rather useless way. For instance, using nonstandard analysis, one can map $f \in C^\infty({\bf R})$ to the nonstandard real number $\int_0^N f(x)\ dx \in {}^* {\bf R}$ for some fixed unbounded real number $N$, and this will be a perfectly well defined additive homomorphism. If one quotients out the infinitesimals $o({\bf R}) := \{ x \in {}^* {\bf R}: x = o(1) \}$ from ${}^* {\bf R}$, one in fact gets a real-linear map that obeys property (1) (if one identifies ${\bf R}$ with a subspace of ${}^* {\bf R}/o({\bf R})$ in the usual manner), but not (2) or (3). But I'm not sure one can do anything particularly interesting with this construction.

  • $\begingroup$ Thanks! What if I just want this $T$ be defined on just a subspace? 1 is translational invariant, that certainly gives some trouble. ($\sin x$ seems to give similar trouble if one translate it by $\pi$.) $\endgroup$
    – 36min
    Commented Dec 9, 2012 at 0:14
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    $\begingroup$ I would like to know your opinion on my answer. Particularly, from the algebraic point of view. $\endgroup$
    – Anixx
    Commented Sep 30, 2019 at 8:38

I don't see why you want $A$ to be an algebra, since the integral of 1 doesn't seem like a reasonable unit. Did you want some compatibility with higher dimensional integrals using the Fubini theorem? Otherwise, if you follow Kähler's lead, it seems more natural to expect a real (or complex) vector space.

Let $C^\infty(\mathbb{R})_{int}$ denote the subspace of $C^\infty(\mathbb{R})$ whose elements are integrable on $[0,\infty)$, and let $C^\infty(\mathbb{R})_{int}^0$ denote the codimension one subspace of functions whose integral is zero. Here's a rephrasing of the desired properties of $A$ and $T$:

  1. Linearity of $T$.

  2. The restriction of $T$ to $C^\infty(\mathbb{R})_{int}$ lands in a distinguished subspace $\mathbb{R} \subset A$, and is given by ordinary integration.

  3. Good behavior under the action of the group $\mathbb{R} \rtimes \mathbb{R}^\times_{>0}$ generated by translations and orientation-preserving dilations.

[Edit:] Let $X$ is a space of smooth functions closed under addition by $C^\infty(\mathbb{R})_{int}$, such that $\mathbb{R} \rtimes \mathbb{R}^\times_{>0}$ acts freely on the quotient vector space $X/C^\infty(\mathbb{R})_{int}$. If a universal target $A$ for integration existed, then $X/C^\infty(\mathbb{R})_{int}^0$ should admit an injection to $A$, because your list of conditions specifies no further relations. The problem (as pointed out by Tao) is that there are lots of smooth functions with nontrivial stabilizer in $\mathbb{R} \rtimes \mathbb{R}^\times_{>0}$.

I think a common method to removing such a difficulty is to ignore the requirement that integration be $\mathbb{R} \rtimes \mathbb{R}^\times_{>0}$-equivariant. Then your universal space is just $C^\infty(\mathbb{R})/C^\infty(\mathbb{R})_{int}^0$.

  • $\begingroup$ $A$ and/or $f$ may be non-unital, i.e. the integral of $1$ may be simply an idempotent in $A$, which seems to be compatible with the intuitive equation $\infty^2=\infty$. $\endgroup$ Commented Dec 8, 2012 at 10:58
  • $\begingroup$ Yes, I suppose my protest is only about half of a reason. $\endgroup$
    – S. Carnahan
    Commented Dec 8, 2012 at 14:22
  • $\begingroup$ S. Carnahan: I don't understand your argument at the end. Why are subspaces relevant? Fernando Muro: Isn't the integral of $1$ obviously $0$ by the third axiom? $\endgroup$
    – Will Sawin
    Commented Dec 8, 2012 at 20:15
  • $\begingroup$ @Will: Okay, I've rewritten the end. $\endgroup$
    – S. Carnahan
    Commented Dec 10, 2012 at 10:39
  • $\begingroup$ I gave a comment on to OP which justifies the word algebra provided we let A to be a non commutative algebra $\endgroup$ Commented Oct 8, 2023 at 12:05

Why make $A$ an algebra? Integration is fundamentally a linear operation that has at best a complicated relationship with multiplication on $C^\infty(\mathbb{R})$. And I see that Scott Carnahan has just made the same point in another answer... Scott's the kernel of what I was going to suggest as well: asymptotic growth classes. Let me expand on that.

Let $I_+ \subset C^\infty(\mathbb{R})$ be the ideal of functions that vanish on some neighborhood of $+\infty$ (for definiteness, say, on at least one interval of the form $[a,+\infty)$ with $a\in\mathbb{R}$). Let $A_+ = C^\infty(\mathbb{R})/I_+$. The quotient $A_+$ is an $\mathbb{R}$-algebra whose elements capture the rates of asymptotic growth at $+\infty$. Let $1_+$ be the image of the constant function $1\in C^\infty(\mathbb{R})$ under the quotient map. Define $I_-$, $A_-$ and $1_-$ in the same way, by replacing $+\infty$ with $-\infty$.

Now, let $B=A_+\oplus A_-$ and $N\subset B$ be the linear subspace spanned by the element $1_+\oplus 1_-$. And finally let $A = B/N$, where we are just taking the quotient of linear spaces (the ring property of $B$ ceases to be of importance).

Any smooth function $f\in C^\infty(\mathbb{R})$ has an indefinite integral $f_a(x) = \int_a^x f(y) dy$ that is also in $C^\infty(\mathbb{R})$. Applying the quotient maps above, we end up with an image $[f_a]$ of $f_a$ in the linear space $A$. The fact that the quotient images of constant functions give zero shows that the images indefinite integrals with different base points (say $f_a$ and $f_b$) coincide. I think that letting $T(f) = [f_a]$ will satisfy all the properties that you wanted of an "integration" map.

  • $\begingroup$ I just noticed that you wanted an integral only from $0$ to $+\infty$ and not from $-\infty$ to $+\infty$. But the exact same argument goes through. Except that you can replace $I_-$ by $I_0$, which would be the ideal of all functions vanishing at $0$ and the quotient $A_0 = C^\infty(\mathbb{R})/I_0$ would be isomorphic to $\mathbb{R}$. $\endgroup$ Commented Dec 8, 2012 at 16:50
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    $\begingroup$ This doesn't satisfy the second relation, since there is no reason for $\int 0^x f(y)dy - \int_0^x f(y+a) dy$ to equal a constant. Moreover, it fails to satisfy the first relation because you have defined the integral of every integrable function to be $0$! $\endgroup$
    – Will Sawin
    Commented Dec 8, 2012 at 20:18
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    $\begingroup$ @Will, sorry, somehow I don't see what you mean by your last remark. The image $T(f)$ of an integrable function in $A_+\oplus A_-$ or $A_+\oplus A_0$ is definitely not zero. Then the quotient by the $\mathbb{R}$-span of $1_+\oplus 1_-$ or $1_+\oplus 1_0$ only kills those indefinite integrals that have the same value at both integration limits, i.e. having finite and zero total integral. $\endgroup$ Commented Dec 9, 2012 at 10:21

I am currently working on a similar system. But your properties (2) and (3) would not work and need change.

Instead, the following properties would work much better:

$$\int_a^c f(x) dx=\int_a^b f(x)dx+\int_b^c f(x)dx\tag{1}$$

$$\int_a^b (f(x)+g(x)) dx=\int_a^b f(x)dx+\int_a^b g(x)dx\tag{2}$$

$$\int_a^b c f(x) dx =c \int_a^b f(x) dx\tag{3}$$

$$\int_{-\infty}^0 f(x) dx=\int_0^\infty f(-x) dx\tag{4}$$

where $a,b,c,f(x)$ and $g(x)$ take values from ℝ ∪ {−∞, +∞}.

Thus, any integral

$$T=\int_a^b f(x) dx$$ represents an "extended" number.

The integrals which are regularizable by a stable method like Cesaro or Abel (as opposed to Ramanujan or Dirichlet) are taken to be equal to their regularized sum:

$$\int_0^\infty f(x)\,dx=\lim_{\epsilon\to 0}\int_0^\infty e^{-\epsilon x}f(x) \, dx\tag{6}$$

Two integrals $\int_0^\infty f(x) dx$ and $\int_0^\infty g(x) dx$ are thus equal if

$$\lim_{\epsilon\to 0}\int_0^\infty e^{-\epsilon x}(f(x)-g(x)) \, dx=0$$

We also can equate some divergent series to the integrals:

$$\sum_{k=0}^\infty f(k)=\int_{-1/2}^\infty\sum_{k=0}^\infty\operatorname{rect}(x+k)f(k)dx$$

In our notation we will consider by definition $$\sum_{k=n}^\infty f(k)=\sum_{k=0}^\infty f(k)-\sum_{k=0}^{n-1}f(k)$$

Now, we postulate that the regularized value or the integral or corresponding series represents the regular part of an extended number, while the rest is the irregular part. Among the suitable regularization methods are Cesaro, Abel, Ramanujan, Borel, Dirichlet regulaization and some others (they agree with each other when applicable). We will denote the regularized value of an extended number $w$ as $\operatorname{reg} w$

Particularly, very useful would be the Faulhaber's formula for Ramanujan's summation of analytic functions:

$$\operatorname{reg} \sum _{n=0}^{\infty} f(n)= -\sum_{n=1}^{\infty} \frac{f^{(n-1)} (0)}{n!} B_n \tag{7}$$

We will use the following symbols for the three most key integrals and series:

$$\omega_+=\sum_{k=0}^\infty 1$$

$$\omega_-=\sum_{k=1}^\infty 1=\omega_+-1$$

$$\tau=\int_0^\infty dx=\omega_+-1/2=\omega_-+1/2$$ (this can also be formally interpreted as $\tau=\pi\delta(0)$ due to Fourier transform).

By interpreting formula (7) as a Taylor series, we come to a formula which allows to generalize the analytic functions to extended numbers (at least in the sense of determining the regular part of the result): $$\operatorname{reg} f'(\omega_-+z)= \Delta f(z)\tag{8}$$ and in particular, to the generalizations of powers of our key series: $$\operatorname{reg}\omega_-^n=B_n\tag{9}$$ $$\operatorname{reg}\omega_+^n=B^*_n\tag{9a}$$ where $B^*$ are the second Bernoulli numbers (those which have $B^*_1=1/2$ ). A more general formula reveals the role of the Hurwitz zeta function: $$\operatorname{reg}(\omega_-+x)^a= B_a(x)=-a\zeta(1-a,x)$$

Based on formula (7) we even can derive an expression for a derivative of an analytic function which does not use limits:

$$f'(x)=\operatorname{reg}(f(\omega_++x)-f(\omega_-+x))=\operatorname{reg} \Delta f(\omega_-+x)$$

which works for any regular $x$.

Also, since many series expansions of trigonometric functions use Bernoulli numbers, we can interpret them as regular parts of similar series but involving extended numbers. This way, and using formula (8) we can obtain the following relations:

$$\operatorname{reg}\sin (a\omega_-+x) = \frac{a}{2} \cot \left(\frac{a}{2}\right) \sin x -\frac{a}{2} \cos x$$

$$\operatorname{reg}\cos (a\omega_-+x) = \frac{a}{2} \csc \left(\frac{a}{2}\right) \cos \left(\frac{a}{2}- x \right)$$

$$\operatorname{reg}\ln (\omega_-+z)=\psi(z)$$

$$\operatorname{reg} e^{z\omega_-}=\frac{z}{e^{z}-1}$$

Particularly, $$\operatorname{reg}\sin \omega_-=-1/2;$$ $$\operatorname{reg}\sin \omega_+=1/2;$$ $$\operatorname{reg}\ln \omega_+=-\gamma;$$ $$\operatorname{reg} e^{\omega_-}=\frac{1}{e-1};$$ $$\operatorname{reg} e^{\omega_+}=\frac{e}{e-1}.$$

Another notable thing is the possibility to express trigonometric functions via inverse trigonometric or logarithms:

$$\cot x=\operatorname{reg}\frac1{\pi }\ln \left(\frac{\omega _+-\frac{x}{\pi }}{\omega _-+\frac{x}{\pi }}\right) = \operatorname{reg}\frac2z \cos (2x\omega_\pm)$$

$$\tan x=\operatorname{reg} \frac1\pi\ln \left(\frac{\tau +\frac{x}{\pi }}{\tau -\frac{x}{\pi }}\right)$$

$$\coth x=\operatorname{reg}\frac{1}{\pi} \operatorname{arccoth}\left(\frac{\pi \omega _+}{x}\right)+\frac1x=\operatorname{reg}\frac1x \cosh (2 x\omega_\pm)$$

Following from Faulhaber's formula (for $ n\ge0 $ ),

$$ \int_0^\infty x^n dx=\frac{\left(\tau +\frac{1}{2}\right)^{n+2}-\left(\tau -\frac{1}{2}\right)^{n+2}}{(n+1)(n+2)}=\frac{\omega _+^{n+2}-\omega _-^{n+2}}{(n+1)(n+2)} $$

Interpreting Fourier transform formally, for even $ n $ we also have

$$ \int_0^\infty x^n dx=i^n\pi\delta^{(n)}(0) $$

For $ n>1 $

$$ \int_0^\infty \frac1{x^n} dx=\frac1{(n-1)!}\int_0^\infty x^{n-2} dx=\frac{\omega _+^{n}-\omega _-^{n}}{(n-1)n!} $$

Particularly, $$\int_0^\infty 1 dx =\tau$$ $$\int_0^\infty x dx=\frac{\tau^2}2+\frac1{24}$$ $$\int_0^\infty x^2 dx=\frac{\tau^3}3+\frac{\tau}{12}$$

We introduce generalized limits in the following way:

$$ \operatorname{gen}\lim_{x\to u^+}f(x)=f(a)-\int_u^a f'(x)dx $$

where $a>u$ and

$$\operatorname{gen}\lim_{x\to u^-}f(x)=f(a)+\int_a^u f'(x)dx$$ where $a<u$

This value can serve as a measure of the growth rate of a function.

Particularly, for $ n\ge0 $

$$ \operatorname{gen}\lim_{x\to\infty}x^n=\frac{\omega _+^{n+1}-\omega _-^{n+1}}{n+1} $$

For odd $ n $ ,

$$ \operatorname{gen}\lim_{x\to\infty}x^n=i^{n-1}\pi n\delta^{(n-1)}(0) $$


$$ \operatorname{gen}\lim_{x\to 0^+} \frac1{x^n}=\frac{\omega _+^{n+1}-\omega _-^{n+1}}{(n+1)!} $$

Following Urs Graf, p.36, for odd $ n $ ,

$$ \operatorname{gen}\lim_{x\to 0^+} \frac1{x^n}=\frac{i^{n-1}\pi\delta^{(n-1)}(0)}{(n-1)!} $$

For instance,

$$\int_{0^+}^\infty \frac1{x^2}=\tau$$

This, combined with the delta function form of $\tau=\pi\delta(0)$ can explain why logarithm has its imaginary part represented as a step function (which is integral of delta function).

The Cauchy principal value of an analytic function at a pole corresponds to the regular part of its generalized limits at the pole, while the order of the pole corresponds to the order of polynomial of $\tau$ of the generalized limit. Thus,



$$\operatorname{gen}\lim_{x\to{-2}^\pm}\Gamma(x)=\frac{3}{4}-\frac{\gamma }{2}\pm\frac\tau 2$$

$$\operatorname{gen}\lim_{x\to{-3}^\pm}\Gamma(x)=\frac{\gamma }{6}-\frac{11}{36}\mp\frac\tau 6$$


For Bernoulli polynomials we have these noticeable formulas:



And more generally,

$$(\omega_-+a)^n=B_n(a)+n\int_0^\infty B_{n-1}(x+a)dx$$

Moreover, generalizing the above formulas we can write down the general formula for conversion of divergent integral representation to omega-tau representation:

$$\int_0^{\infty } f(x) \, dx=\operatorname{reg}\int_0^{\infty } f(x) \, dx+\int _{\omega_-}^{\omega_+} \int _0^x f(t) dtdx$$

The integral from $\omega_-$ to $\omega_+$ should be understood as the difference of the antiderivative on those values.

Here is a table of some extended numbers and their representations in different forms:

$$ \begin{array}{cccccc} \text{Delta form} & \text{In terms of } \tau, \omega_+,\omega_- & \text{Regular part} & \text{Integral or series form} & \text{Generalized limit form} \\ \pi \delta (0) & \tau & 0 & \int_0^{\infty } \, dx;\int_0^{\infty } \frac{1}{x^2} \, dx & \operatorname{gen}\lim_{x\to\infty}x;\operatorname{gen}\lim_{x\to0^+}\frac1x \\ \pi \delta (0)-\frac{1}{2} & \omega _-;\tau-\frac{1}{2} & -\frac{1}{2} & \sum _{k=1}^{\infty } 1 & \operatorname{gen}\lim_{x\to\infty} (x-1/2) \\ \pi \delta (0)+\frac{1}{2} & \omega _+;\tau+\frac{1}{2} & \frac{1}{2} & \sum _{k=0}^{\infty } 1 & \operatorname{gen}\lim_{x\to\infty} (x+1/2) \\ 2 \pi \delta (i) & e^{\omega_+}-e^{\omega_-}-1 & 0 & \int_{-\infty }^{\infty } e^x \, dx & \operatorname{gen}\lim_{x\to\infty} e^x \\ & \frac{\tau ^2}{2}+\frac{1}{24};\frac{\omega_+^3-\omega_-^3}6 & 0 & \int_0^{\infty} x \, dx;\int_0^\infty \frac2{x^3}dx & \operatorname{gen}\lim_{x\to\infty}\frac{x^2}2;\operatorname{gen}\lim_{x\to{0^+}} \frac1{x^2}\\ & \frac{\tau ^2}{2}-\frac{1}{24} & -\frac1{12} & \sum _{k=0}^{\infty } k & \operatorname{gen}\lim_{x\to\infty} \left(\frac{x^2}2-\frac1{12}\right) \\ -\pi \delta''(0) &\frac {\tau^3}3 +\frac\tau{12};\frac{\omega_+^4-\omega_-^4}{12}& 0 & \int_0^\infty x^2dx;\int_0^\infty\frac6{x^4}dx&\operatorname{gen}\lim_{x\to\infty}\frac{x^3}3;\operatorname{gen}\lim_{x\to{0^+}} \frac2{x^3}\\ \pi^2\delta(0)^2-\pi\delta(0)+1/4&\omega_-^2&\frac16&2 \int_0^{\infty } \left(x-\frac{1}{2}\right) \, dx+\frac{1}{6}&\operatorname{gen}\lim_{x\to\infty}B_2(x)\\ \pi^2\delta(0)^2+\pi\delta(0)+1/4&\omega_+^2&\frac16&2 \int_0^{\infty } \left(x+\frac{1}{2}\right) \, dx+\frac{1}{6}&\operatorname{gen}\lim_{x\to\infty}B_2(x+1)\\ \pi^2\delta(0)^2&\tau^2&-\frac1{12}&\int_{-\infty}^{\infty } |x| \, dx-\frac{1}{12}&\operatorname{gen}\lim_{x\to\infty}B_2(x+1/2)\\ &\ln \omega_++\gamma&0&\int_1^\infty \frac{dx}x;\sum_{k=1}^\infty \frac1x -\gamma&\operatorname{gen}\lim_{x\to\infty}\ln x\\ -\pi\delta''(0)-\frac14 \pi\delta(0);\pi^3\delta(0)^3&\tau^3&0&\int_0^\infty \left(3x^2-\frac1{4}\right)dx&\operatorname{gen}\lim_{x\to\infty}B_3(x+1/2)\\ \frac{2\pi\delta(i)+1}{e-1}&e^{\omega_-}&\frac1{e-1}&\frac1{e-1}+\frac1{e-1}\int_{-\infty}^\infty e^x dx&\operatorname{gen}\lim_{x\to\infty} \frac{e^x+1}{e-1}\\ \frac{2\pi\delta(i)+1}{1-e^{-1}}&e^{\omega_+}&\frac1{1-e^{-1}}&\frac1{1-e^{-1}}+\frac1{1-e^{-1}}\int_{-\infty}^\infty e^x dx&\operatorname{gen}\lim_{x\to\infty} \frac{e^x+1}{1-e^{-1}}\\ \end{array} $$

The system forms an integral domain which can be extended to form a field. Still, there are many white areas, particularly, there is no straightforward way to transform a divergent integral into an expression in terms of $\omega_\pm$ and vice versa. There is no formula to construct an integral representing a product of two integrals, etc.

  • 5
    $\begingroup$ 32nd version of this answer! $\endgroup$ Commented Nov 20, 2019 at 21:34
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    $\begingroup$ @GerryMyerson Now up to 33. To the answerer, I think the point is that these endless revisions are not really appropriate - this is becoming more a diary of personal research than an answer to the specific question. $\endgroup$ Commented Nov 30, 2019 at 19:00
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    $\begingroup$ @Noah Schweber I am inclined to fix mistakes rather than allow them persist. $\endgroup$
    – Anixx
    Commented Nov 30, 2019 at 19:05
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    $\begingroup$ @Anixx But you've been doing a lot more than just fixing mistakes. Seriously, 33 edits is just too many - keep in mind that each time you edit, this question jumps back onto the front page of the site. $\endgroup$ Commented Nov 30, 2019 at 19:08
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    $\begingroup$ @Anixx Please don't. $\endgroup$
    – R.P.
    Commented Aug 17, 2020 at 17:01

EDIT: Let $g(x)$ be any function that is smooth at $0$, meaning all its derivatives vanish at $0$. Let $f(x)$ satisfy $f(x)=g(x)+f(x-1)$ for $x>0$ and $f(x)=0$ for $x\leq 0$. Clearly $f$ is smooth. Then

$\int_0^{\infty} g(x) = \int_0^\infty f(x) - \int_0^\infty f(x-1) = \int_0^\infty f(x) - \int_0^\infty f(x)=0$

So the second axiom makes any function which is smooth at $0$ vanish. This contradicts the first axiom.

You can throw out all the axioms but the first, add an obvious algebra structure, and still run into trouble.

Using integration by parts, we can see that $\int u dv + \int v du = [uv]_0^{\infty}$. If $u(0)=v(0)=0$, then this is equal to $[u]_0^\infty [v]_0^\infty$.

By $[u]_0^\infty$, I mean $\int_0^\infty du$.

If you accept that equality, and you want the integral of anything that is actually integrable with integral $0$ to be $0$, then you've got a problem. Take $v$ to be a function that decreases to $0$ at $\infty$ but is never $0$, and take $u=f/v$ for any function $f$ (that vanishes to second order at $0$). Then

$\int df= \int udv+\int vdu= [u]_0^{\infty} [v]_{0}^\infty = [u]_0^\infty 0 = 0$

Since the local conditions on $f$ at $0$ clearly don't matter, the integral of any function is zero.

It's clear why this happens. Using $\int_0^\infty f(x) dx = \lim_{y\to \infty} \int_0^y f(x) dx$, we see that integration is exactly the same problem as taking limits. The only way to take the limit of an arbitrary function is an ultrafilter. But ultrafilters usually take decreasing functions to (invertible) infinitesimals, not $0$.

So this is a good explanation of why you're forced to give up the algebra structure, I think - it prevents you from sending functions that integrate to $0$ to $0$.

  • 3
    $\begingroup$ I think the more common term is "flat at 0" instead of "smooth at 0". $\endgroup$
    – S. Carnahan
    Commented Dec 10, 2012 at 4:32

Since specific pre-determined axioms might have problems, to formalize the problem to give a reasonable value to integrals $\int_0^\infty f$ (with real $f$) where each $F(x)=\int_0^x f$ reasonably exists for each finite $x$, I suggest to look at the universal compactification of $[0,\infty)$ taking into account the amenability of this additive semigroup (in the same way as amenability of the additive semigroup of natural integers gives Banach limits of bounded sequences). [The standard reference for rings of continuous functions is Gillman - Jerison; I will not give references for amenability, only note Wagon's book that relates in detail failure of amenability to existence of paradoxical decompositions like Banach - Tarski - Hausdorff]

The above functions $x\mapsto F(x)$ are continuous, hence when bounded (oscillating integrals) have a unique real continuous extension to the universal compactification $\beta[0,\infty)$; as value at infinity the "universal value" is then an element, call it $F(\infty)$, of the algebra $C(X)$ of continuous real valued functions on the remainder $X=\beta[0,\infty)\setminus[0,\infty)$. $F(\infty)$ has a unique real value, i.e. $F$ extends by continuity to the one point compactification of $[0,\infty)$, iff it is a constant function in $C(X)$. Commutativity, hence amenability, of $[0,\infty)$ gives (independence of $F(x)$ from the value of $f$ in fixed finite segments $[0,x]$ being automatic) that we can also chose a "almost reasonable" single real value instead of a family $F(\infty)$ of such values, by chosing a suitable positive measure $\mu$ on the remainder $X$ and integrating: $\int_X F(\infty)d\mu$ (choosing a Dirac's delta for $\mu$ means chosing one of the points of the remanider $X$ i.e. choosing a ultrafilter of co-zero sets in the original space).

This is perhaps the best that can be done for associating reasonable values to boundedly oscillating integrals. When $F$ is not bounded, it is still continuous; by Gelfand - Kolmogorov, the maximal spectrum of the ring $R$ of (possibly unbounded) real continuous functions is the same (universal compactification $\beta$) as that of the ring of bounded real continuous functions, and $F(\infty)$ is then again definable as function on the remainder $X$, but this time its values in each point at infinity (i.e. maximal ideal $m$ of the ring $R$) need not real numbers, but elements of a noarchimedean real-closed extension field $R/m$ of the real numbers (that field depends upon $m$, but a common extension exists, by the amalgamation property of real closed fields: Hodges, model theory, pp. 384 - 386). So one has reached nonstandard analysis and "orders of infinity" like in the previous answers & comments. What changes for general $F$ when comparing with the subcase of bounded $F$ is that this time no obvious averaging process exists to obtain a single hyper-real number instead of a family $F(\infty)$ of such numbers. [Or, more exactly, I know no theory of measure and integration with hyper-real valued functions and measure that would easily integrate all $F(\infty)$ in the same way as the standard theory works for the real, archimedean, case]

The preceding considerations are by construction translation invariant, but the action of the group of dilations (last axiom) is not considered. The group generated by translations and dilations (i.e. the group of affine transformations of the real line) is the semidirect product of two abelian groups, hence soluble hence amenable. So perhaps it might be possible to obtain a kind of suitable invariance / covariance also for dilations, despite the fact that the explicitly noted axioms have problems.


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