Is there a formula or algorithm to remove infinitesimal and oscillating parts from an expression while keeping finite and infinite ones? [closed]

Question

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?

Answer 1

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