# A Rational Solution for a Rational Function

$\int_{-1}^{1} \frac{1}{x} \, dx$

Calculus tells us, with a great deal of finesse, that the above integral has no solution; logic, however, tells us it does. Allow me to explain.

First, let us note that $0$ is not within the domain of $x$ -- $[ \text{i.e.} \: 0 \notin X ]$.

Let us also note that for every $x$ within the interval $[ -1, \, 0 )$, there is a corollary "inverse" $x$ in the interval $( 0, \, 1 ]$ -- $[ \text{i.e.} \: \forall i \in \{ [ -1, 0) \}, \: [ \exists j \in \{ (0, 1 ] \} \, | \, ( i + j ) = 0 ] \; \text{and vice versa} ]$

Thus, even without "arithmetising infinity", we can see that the integrated sum of each side of the equation does, indeed, cancel out.

Unless I have missed something? Note by Joshua Nesseth
4 years, 5 months ago

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Yes. You missed out on understanding why we can only arbitrarily rearrange a series if and only if the series is absolutely convergent.

Once you understand that, notice that your argument about "pair up the inverses" is basically doing an arbitrary rearrangement. Unfortunately, because your series isn't absolutely convergent, this step is disallowed.

Staff - 4 years, 5 months ago

Are we not already assuming an infinite set of input values for any non-discrete variable integration?

Moreover, as long as the (even infinite) set is well-ordered, we are not performing "arbitrary" rearrangement, are we?

Logically, the pairing is not arbitrary. If we have "inverse values" for each value over an (infinite) set in another (pair-wise infinite) set, and nothing in either set does not have a corollary value in the other, why could we not pair each up?

[I do understand given non-converngence, the infinite sum cannot be used as a method of proof--but it does not preclude a different form of proof from a set-theoretic perspective, does it? (That's an honest question...)]

- 4 years, 5 months ago

You're at the area crossing from "let's wave our hands and sweep all the details under the rug" and getting into "let's nail down the nitty gritty details of why these things are true". This is where understanding the concepts and its details gets important.

Let me first explain how the "arbitrary rearrangement" comes in. Consider the intervals $A_n = [ - \frac{1}{2^n} , - \frac{ 1}{ 2^{n+1} } ]$ and $B_n = [ \frac{1}{2^{n+1} } , \frac{ 1}{ 2^n } ]$. Let the intergral of $\frac{1}{x}$ over these intervals be $a_n, b_n$ respectively.

Then, the (naive) claim is that the integral is equal to the sum of the series:

$a_0 + a_1 + a_2 \ldots + "a_{\infty}" + "b_{\infty}" + \ldots + b_2 + b_1 + b_0 .$

What you are doing is rearranging the terms "arbitrarily" (more accruately, the phrase arbitrarily refers to moving the terms massively around, as opposed to a random reordering) into the form

$a_0 + b_0 + a_1 + b_1 + a_2 + b_2 + \ldots + "a_{\infty}" + "b_{\infty}"$

Reading through absolutely convergent, a necessary and sufficient condition for the arbitrary rearrangement to converge to the same limit, is that the sequence is absolutely convergent. However, because this sequence is not true, the limit does not stay the same. In fact, we could rearrange the terms in such a way to obtain any value of the limit that we desire.

Recall that when dealing with improper integrals, we have to take the limit of the definite integral as the interval approaches its desired size.

In the case of $\int_0^\infty f(x) \, dx$, we want to ensure that the limit $\lim_{a \rightarrow \infty} \int_0^a f(x) \, dx$ exists. For example, $\int_0^\infty \cos x \, dx$ does not exist, because the underlying sequence osscilates.

In the case where $f(x) \rightarrow \infty$, then we have to take limits from the left and right. Using $\int_{-1}^1 \frac{ 1}{x} \, dx$ as an example, we have to consider the 2-variable limit

$\lim_{(a,b) \rightarrow (0^-, 0^+ ) } \int_{-1} ^a \frac{1}{x} \, dx + \int_{b}^1 \frac{1}{x} \, dx .$

Because this limit over all possible paths towards $(0^-, 0^+)$ doesn't exist, hence the improper limit does not exist.

Staff - 4 years, 5 months ago

The refutation you must provide (and I'm assuming you can, only saying that you haven't) is one based in set theory--not in the principles of calculus. As I have acknowledged quite extensively, I don't deny the integral's non-existence under the rules of integration--I only assert it's conceptual existence independent of the current confines of mathematical integration (perhaps a flaw, perhaps simply a requisite short-coming).

I understand why, in specific terms of an infinite sequence, the operations are not valid--one cannot simply "cancel out" an infinite sum with another "inverse" infinite sum. That, however, is not what I am claiming; nor am i claiming to have reached the solution of the problem through integration (despite using the integral as a starting-point...though allow me to explain).

My argument is strictly from a set-theoretic perspective. I am arguing that in a(n albeit infinite,) well-ordered set with a defined, unique "inverse" counterpart for each (infinitely numerous) member of that set, such that the converse is also true (i.e. vice versa), there is a unique, logical "pairing" for each member of each set; this requires no addition whatsoever.

My claims of this being a "solution" to the integral is, of course, not exactly valid--it (as I mention in the original post) is wholly outside the purview of integral calculus. My point, however, is that such a solution SHOULD exist--meaning that, in this case, we can show that there is a good, rigorous reason there SHOULD be an answer to the question, even if only in concept.

[This is part of a larger position I have on the application of mathematics in general--but that can wait...for now.]

- 4 years, 5 months ago

Again, I may be overstepping my station (as I often do), but while I understand the limits of limits (sorry, I had to) and integral calculus as a whole, my argument is that when we examine the integral as a process, there is a way to guarantee the answer is zero--even if we have to delve into set-theory for a proof.

- 4 years, 5 months ago