A Rational Solution for a Rational Function

111xdx \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 0 is not within the domain of x x -- [i.e.0X] [ \text{i.e.} \: 0 \notin X ] .

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

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}


Sort by:

Top Newest

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.

Calvin Lin Staff - 4 years, 5 months ago

Log in to reply

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...)]

Joshua Nesseth - 4 years, 5 months ago

Log in to reply

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 An=[12n,12n+1] A_n = [ - \frac{1}{2^n} , - \frac{ 1}{ 2^{n+1} } ] and Bn=[12n+1,12n] B_n = [ \frac{1}{2^{n+1} } , \frac{ 1}{ 2^n } ] . Let the intergral of 1x \frac{1}{x} over these intervals be an,bna_n, b_n respectively.

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

a0+a1+a2+"a"+"b"++b2+b1+b0. 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

a0+b0+a1+b1+a2+b2++"a"+"b" 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 0f(x)dx \int_0^\infty f(x) \, dx, we want to ensure that the limit lima0af(x)dx \lim_{a \rightarrow \infty} \int_0^a f(x) \, dx exists. For example, 0cosxdx \int_0^\infty \cos x \, dx does not exist, because the underlying sequence osscilates.

In the case where f(x) f(x) \rightarrow \infty , then we have to take limits from the left and right. Using 111xdx \int_{-1}^1 \frac{ 1}{x} \, dx as an example, we have to consider the 2-variable limit

lim(a,b)(0,0+)1a1xdx+b11xdx. \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+) (0^-, 0^+) doesn't exist, hence the improper limit does not exist.

Calvin Lin Staff - 4 years, 5 months ago

Log in to reply

@Calvin Lin 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.]

Joshua Nesseth - 4 years, 5 months ago

Log in to reply

@Calvin Lin 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.

Joshua Nesseth - 4 years, 5 months ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...