-2

Below, we interpret divergent integrals as germs of partial integrals at infinity:

$$\int_0^\infty f(x) dx=\operatorname{bigpart} \int_0^\omega f(x) dx$$

where $\operatorname{bigpart}$ means taking finite and infinite parts of $\int_0^\omega f(x) dx$ at $\omega\to\infty$ while throwing away infinitesimal and oscillating (with zero average) parts. The question is, how can we define this operator?

For instance,

$$\int_0^\omega 1 dx=\omega.$$

It is infinite, so our desired result is $\int_0^\infty 1 dx=\omega$. Thus, $\omega$ plays the role of an infinite constant.

$$\int_0^{\omega } \exp (x) \sin (2 x) \, dx=\frac{2}{5}-\frac{2}{5} e^{\omega } (\cos (2 \omega )-\sin (\omega ) \cos (\omega )).$$

Here, $2/5$ is finite part. The term $-\frac{2}{5} e^{\omega } (\cos (2 \omega )-\sin (\omega ) \cos (\omega ))$ is oscillating with zero average, so should be taken to be equal to zero. So, the desired result is $2/5$.

$$\int_0^{\omega } \cos ^2(x) \, dx=\frac{\omega }{2}+\frac{1}{4} \sin (2 \omega ).$$

The part $\frac{1}{4} \sin (2 \omega )$ is oscillating with zero average, so the desired result is the infinite part $\frac{\omega }{2}$.

$$\int_0^{\omega } \exp (\log (x)+x) \, dx=e^{\omega } (\omega -1)+1.$$

Here we have infinite and finite parts, so the desired result should be kept intact: $\int_0^\infty \exp (\log (x)+x) dx=e^{\omega } (\omega -1)+1$.

On the other hand,

$$\int_0^{\omega } \exp (\log (x)-x) \, dx=e^{-\omega } (-\omega -1)+1.$$

The term $e^{-\omega } (-\omega -1)$ is infinitesimal, so we throw it away, and our desired result is $\int_0^{\infty } \exp (\log (x)-x) \, dx=1$.

Is there a way to define such function $\operatorname{bigpart}$ consistently and automatize the process in a CAS system?

LSpice
  • 11,423
Anixx
  • 9,302
  • This seems to be a question about a particular number system, that includes (I suppose) the standard real numbers, as well as infinite, infinitesimal, and oscillating (?) numbers. Could you give a reference to the definition of this system? Also, the question about whether the process can be automatised in a CAS seems to be a question about CAS's (maybe for Theoretical CSSE, or MathematicaSE?), not about mathematics per se. – LSpice Jul 29 '22 at 19:10
  • @LSpice the question talks about behavior of expressions of $\omega$ as $\omega$ tends to infinity: it can be infinitesimal (decreasing by magnitude), infinite (growing by magnitude) or oscillating (nonzero but with zero average), as well as a sum of the above. – Anixx Jul 29 '22 at 19:12

1 Answers1

2

This isn't an answer, just a long comment.

If this is from an established field, and I'd guess it is, that needs to be part of the question. Not knowing one, I will blindly sally forth because I am intrigued.

The use of $\omega$ as a variable is discordant. Whatever you are doing comes to the same thing (I think) if reformulated as:

Given a function $f(x)$ , give a meaning to $$\lim_{x \rightarrow \infty}\int_0^xf(t)dt.$$

In fact, integrals seem irrelevant here. You are not concerned with the art of evaluating integrals but rather the behavior of the resulting function expressions as $x$ goes to $\infty$.

I'll take your comment as: Given $F(x)$ write it as $M(X)+V(x)+O(x)$ where $M$ is (eventually) monotonic, $V(x)=o(x)$ and $O(x)$ is "oscillating".

In pursuit of oscillating:

What would you say is $\lim_{x \rightarrow \infty}\sin(\ln(x+1))$ or, equivalently $\lim_{x\rightarrow \infty} \int_0^x\frac{\cos(\ln(t+1))}{t+1}\,dt$?

What about $\lim_{x \rightarrow \infty}(x+1)\sin(\ln(x+1))$ or, equivalently $\lim_{x\rightarrow \infty} \int_0^x\sin(\ln(t+1))+\cos(\ln(t+1))\,dt$?

LSpice
  • 11,423
Aaron Meyerowitz
  • 30,015
  • Actually, there should be 4 terms, finite, infinite, infinitesimal and oscillating (with zero average). We need to keep only first two. – Anixx Jul 29 '22 at 22:22
  • 4
    Using $o(x)$ to mean "goes to $0$ as $x \to 0$" and $O(x)$ to mean "oscillatory" might not be a great idea. – LSpice Jul 29 '22 at 22:22
  • 2
    @Anixx, I am a little confused about your infinitesimal piece (what I think this answer takes to be the piece $V(x)$). It can be added on to an infinite (i.e., increasing without bound) piece, which will remain infinite. So how should one know to separate it out? For example, how should I know (as I assume I should) to keep $x$ and discard $\frac1 x$ in the expression $x + \frac1 x$? – LSpice Jul 29 '22 at 22:26
  • @LSpice yes, I am also a bit confused about this, so that's why my question. But I can produce a lot of examples of what results I want. – Anixx Jul 29 '22 at 22:31
  • @LSpice actually, that's the main problem in this question. – Anixx Jul 29 '22 at 22:32
  • @LSpice if the expression is a power series (including Laurent series), then the answer is obvious, not so obvious in other cases. – Anixx Jul 29 '22 at 22:36
  • @LSpice many functions can be represented as Laurent series though. – Anixx Jul 29 '22 at 22:37
  • A representation by a Laurent series requires a centre point in order to distinguish the non-negative- from the negative-valuation terms. I guess you are taking $0$ as the implicit choice of centre? – LSpice Jul 29 '22 at 23:03
  • @LSpice yes. A power series of $\omega$. $\sum \omega^k$ So, the representation for functions representable as Laurent series is clear, but, can it be generalized to other classes? – Anixx Jul 29 '22 at 23:05
  • The Laurent (and Taylor) series of $(x+1)\ln(x+1)$ is $x +\frac{1}{2} x^{2}-\frac{1}{3} x^{3}+\frac{1}{6} x^{4}-\frac{1}{12} x^{5}+\frac{1}{24} x^{6}-\frac{5}{252} x^{7}+\frac{1}{126} x^{8}-\frac{23}{18144} x^{9}+\cdots.$ What would you want to do with that? – Aaron Meyerowitz Jul 30 '22 at 01:27
  • That's a difficult question, really. For instance, whether $\ln \omega=\ln (\omega+1)$. I meant, if the series is not changing sign, it is easy, if it does, it is difficult. – Anixx Jul 30 '22 at 07:51
  • I think, we should convert $\ln (\omega+1)\to\ln (\omega)$ because their difference is infinitesimal. Then it comes that $(\omega+1)\ln(\omega+1)\to(\omega+1)\ln \omega$. This is the desired result, I think. – Anixx Jul 30 '22 at 07:56
  • $\lim_{x \rightarrow \infty}\sin(\ln(x+1))\to0$, definitely. – Anixx Jul 30 '22 at 08:00
  • 1
    Because it oscillates, although the running average does not go to zero. So next $\lim_{x \rightarrow \infty}x\sin(ln(x))?$ – Aaron Meyerowitz Jul 30 '22 at 22:28